{"id":20351,"date":"2020-01-20T17:27:01","date_gmt":"2020-01-20T11:57:01","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20351"},"modified":"2021-07-02T15:28:14","modified_gmt":"2021-07-02T09:58:14","slug":"2nd-puc-basic-maths-question-bank-chapter-15-ex-15-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-15-ex-15-2\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.2"},"content":{"rendered":"

Students can Download Basic Maths Exercise 15.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.2<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Circles Ex 15.2 One Mark Questions and Answers<\/strong><\/p>\n

Question 1.
\nFind the centre and the radius of the circle.
\n(a) x2<\/sup> + y2<\/sup> – 4x – y – 5 = 0
\n(b) 3x2<\/sup> + 3y2<\/sup> – 6x – 12y – 2 = 0
\n(c) (x – 2) (x – 4) + (y – 1) (y – 3) = 0
\n(d) x2<\/sup> + y2<\/sup> – 2x cos\u03b1 – 2y sin\u03b1 = 1
\nAnswer:
\n(a) x2<\/sup> + y2<\/sup> – 4x – y – 5 = 0 Comparing with
\nx2<\/sup> + y2<\/sup> + 2gx + 2fy + C = 0 We get
\n2g = – 4, 2f = – 1 \u21d2 g = -2, f = \\(-\\frac{1}{2}\\)
\n\u2234 C=(-8,-1) = (2, \\(\\frac{1}{2}\\))
\n\\(r=\\sqrt{g^{2}+f^{2}-C} ; \\quad r=\\sqrt{(-2)^{2}+\\left(-\\frac{1}{2}\\right)^{2}-(-5)}=\\sqrt{4+\\frac{1}{4}+5}=\\sqrt{\\frac{16+1+20}{4}}=\\frac{\\sqrt{37}}{2} \\text { units }\\)<\/p>\n

(b) Given 3x2<\/sup> + 3y2<\/sup> – 6x – 12y – 2 = 0, divide by 3
\nx2<\/sup> + y2<\/sup> – 2x – 4y – \\(\\frac{2}{3}\\) = 0
\nHere g = -1, f = -2, & c = \\(-\\frac{2}{3}\\)
\n\u2234Centre = (-g, -f) = (1, 2) & r =\\(\\sqrt{g^{2}+f^{2}-C}\\)
\n\\(=\\sqrt{(-1)^{2}+(-2)^{2}-\\left(-\\frac{2}{3}\\right)}\\)
\nr = \\(\\sqrt{1-4-\\frac{2}{3}}=\\sqrt{\\frac{17}{3}}\\) units.<\/p>\n

\"KSEEB<\/p>\n

(c) Given (x – 2) (x – 4) + (y – 1) (y – 3) = 0
\n\u21d2 x2<\/sup> – 6x + 8 + y2<\/sup> – 4y + 3 = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x – 4y + 11 = 0
\nHere g = -3, f= -2, C = 11 & \u2234 Centre = (3,2).
\nAnd r = \\(\\sqrt{9+4-11}=\\sqrt{2}\\) units<\/p>\n

(d) Given x2<\/sup> + y2<\/sup> – 2x cos \u03b1 – 2y sin \u03b1 – 1 = 0
\ng = – cos\u03b1, f = – sin\u03b1, c = -1
\n\u2234 Centre = (cos\u03b1, sin\u03b1) & r = \\(\\sqrt{\\cos ^{2} \\alpha+\\sin ^{2} \\alpha+1}=\\sqrt{1+1}=\\sqrt{2}\\)<\/p>\n

Question 2.
\nIf the radius of the circle x2<\/sup> + y2<\/sup> + 4x – 2y- k = 0 is 4 units find k.
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> + 4x – 2y – k = 0 & r = 4 units, K = ?
\nHere g = 2, f = -1 and C = – k
\n\u2234 Centre = (-2, 1) & r = \\(\\sqrt{g^{2}+f^{2}-c}\\)
\n4 = \\(\\sqrt{(2)^{2}+(-1)^{2}-(-k)}\\) \u21d2 4 = \\(\\sqrt{4+1+k}\\) S.B.S
\n16 = 5 + k \u21d2 k = 16 – 5 = 11<\/p>\n

Question 3.
\nFind the other end of the diameter, if one of the diameter of the circle.
\n(a) x2<\/sup> + y2<\/sup> = 25 is (5,0)
\n(b) x2<\/sup> + y2<\/sup> + 4x – 6y – 12 = 0 is (-5, -1)
\n(c) x2<\/sup> + y2<\/sup> – 6x + 2y = 31 is (7,4)
\nAnswer:
\n(a) Given x2<\/sup> + y2<\/sup> = 25, A = (5,0) Let (x2<\/sub> y2<\/sub>) = ?
\nHere C =(0,0) & We know that centre is the midpoint of the diameter.
\n\"2nd
\n\u21d2 5 + x2<\/sub> = 0 0 + y2<\/sub> = 0
\n\u21d2 x2<\/sub> = -5 y2<\/sub> = 0
\n\u2234 The other end of the diameter B = (-5,0).<\/p>\n

(b) Given x2<\/sup> + y2<\/sup> + 4x – 6y – 12 = 0,
\n\"2nd
\nHere C = (-2,3)
\n\u2234 By mid point formula we have
\n\"2nd
\n\u21d2 x2<\/sub> = -4 + 5 = 1 & y2<\/sub> = 6 + 1 = 7
\n\u2234 The other end of the diameter B = (1,7)<\/p>\n

\"KSEEB<\/p>\n

(c) Given x2<\/sup> + y2<\/sup> – 6x + 2y – 31 = 0, A = (7,4), B = (x2<\/sub>, y2<\/sub>) = ?
\nHere C = (3,-1)
\n\"2nd
\n\u21d2 7 + x2<\/sub> = 6 & 4 + y2<\/sub> = -2
\n\u21d2 x2<\/sub> = -1 y2<\/sub> = -6
\n\u2234 Other end of the diameter B = (-1,-6).<\/p>\n

Question 4.
\nIf(a, b) and (-5, 1) are the two end points of diameter of the circle x2<\/sup> + y2<\/sup> + 4x – 4y = 2. Find the value of a and b.
\nAnswer:
\nGiven (x1<\/sub> y1<\/sub>) = (a, b) (x2<\/sub>, y2<\/sub>) = (-5, 1) & Center of the given circle = (-2,2).
\nEquation of the circle when ends of diameter is given is
\n(x – x1<\/sub>) (x – x2<\/sub>) + (y – y1<\/sub>) (y – y2<\/sub>) = 0.
\n(x – a) (x + 5) + (y – b) (y – 1) = 0.
\n\u21d2 x2<\/sup> – 5a – ax + 5x + y2<\/sup> – by – y + b = 0.
\nx2<\/sup> + y2<\/sup> + x (5 – a) – y(b + 1) + (b – 5a) = 0 comparing.
\nThis by given circle x2<\/sup> + y3<\/sup> + 4x – 4y – 2 = 0 we get
\n5 – a = 4, b + 1 = 4
\n\u21d2 5 – 4 = a, b = 4 – 1
\n\u2234 a = 1 & b = 3.<\/p>\n

Question 5.
\nIf x2<\/sup> + y2<\/sup> – 4x – 8y + k = 0 represents a point circle find k
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> – 4x – 8y + k = 0 is a point circle
\n\u21d2 r = 0, g = -2, f = -4 & c = k
\nand \\(\\sqrt{g^{2}+f^{2}-C}=r\\)
\n\\(\\sqrt{4+16-\\mathrm{k}}=0 \\quad=\\sqrt{20-\\mathrm{k}=0}\\) \u21d2 k = 20 S.B.S<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIf x2<\/sup> + y2<\/sup> + ax + 2y + 5 = 0 represent a point circle find a
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> + ax + 2y + 5 = 0 is a point circle.
\n\u21d2 r = 0, g = \\(+\\frac{a}{2}\\) f = 1, C = 5
\nand \\(\\sqrt{g^{2}+f^{2}-C}=r=0\\)
\n\"2nd<\/p>\n

Question 7.
\nIf x2<\/sup> + y2<\/sup> + ax + by = 3 represents a circle with centre at (1, -3), find a and b.
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> + ax + by – 3 = 0 & center = (1, -3), a = ?, b = ?
\nHere \\(\\frac { a }{ 2 }\\) = 1 & \\(\\frac { b }{ 2 }\\) = -3
\n\u21d2 a = 2 & b = – 6.<\/p>\n

Question 8.
\nFind the unit circle concentric with the circle x2<\/sup> + y2<\/sup> – 8x + 4y = 8.
\nAnswer:
\nHere unit circle \u21d2 r= 1
\n& Center of concentric circle x2<\/sup> + y2<\/sup> – 8x + 4y – 8 = 0
\nC = (4, -2) & r =\\(\\sqrt{(-4)^{2}+(2)^{2}-(C)}=\\sqrt{16+4-C}=1\\)
\n\u21d2 20 – C = 19
\n\u21d2 C = 19
\n\u2234 The Equation of the circle is x2<\/sup> + y2<\/sup> – 8x + 4y + 19 = 0.<\/p>\n

\"KSEEB<\/p>\n

Part-B<\/span><\/p>\n

2nd PUC Basic Maths Circles Ex 15.2 Two Marks Question and Answers<\/strong><\/p>\n

Question 1.
\nFind the equation of the circle whose centre is same as the centre of the circle x2<\/sup> + y2<\/sup> + 6x + 2y + 1 = 0, and passing through the point (-2, -3).
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> + 6x + 2y + 1 = 0, P = (-2,-3)
\nCentre = C(-3,-1),
\nLet the equations of the required circle is x2<\/sup> + y2<\/sup> + 6x + 2y + C = 0
\nr = CP = \\(\\sqrt{(-2+3)^{2}+(-3+1)^{2}}=\\sqrt{1^{2}+0}=1\\)
\n\u2234 r = \\(\\sqrt{g^{2}+f^{2}-c}\\)
\nl = \\(\\sqrt{9+1-c}\\) \u21d2 1 = 10 -c \u21d2 c = 9
\n\u2234 The required equation of the circle is
\nx2<\/sup> + y2<\/sup> + 6x + 2y + 9 = 0.<\/p>\n

Question 2.
\nFind the equation of the circle whose centre is same as the centre of the circle x2<\/sup> + y2<\/sup> – 6x + 4y + 9 = 0 and Passsing through the point (-2,3)
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> – 6x + 4y + 9 = 0, & P = (-2,3)
\nCentre = c(3,-2), P = (-2,3)
\nr = cp = \\(\\sqrt{(-2-3)^{2}+(3-(-2))^{2}}=\\sqrt{25+25}=\\sqrt{50}\\)
\nLet equation of the required circle is
\nx2<\/sup> + y2<\/sup> – 6x + 4y + c = 0
\nr = \\(\\sqrt{g^{2}+f^{2}-c}=\\sqrt{9+4}\\) \u2234 \\(\\sqrt{50}=\\sqrt{13-c}\\) \u21d2 50 = 13 – c \u21d2 c = – 37
\n\u2234 The required circle is x2<\/sup> + y2<\/sup> – 6x + 4y – 37 = 0.<\/p>\n

Question 3.
\nFind the equation of the circle whose centre is (-2, 3) and passing through the centre of the circle x2<\/sup> + y2<\/sup> – 6x + 4y+9=0.
\nAnswer:
\nGiven c= (-2,3) & given x2<\/sup> + y2<\/sup> – 6x + 4y + 9 = 0. P = (3,-2)
\nr = cp = \\(\\sqrt{(3+2)^{2}+(-2-3)^{2}}=\\sqrt{25+25}=\\sqrt{50}\\)
\nAlso r = \\(\\sqrt{g^{2}+f^{2}-c}\\)
\n\\(\\sqrt{50}=\\sqrt{4+9-c}\\) \u21d2 50 = 13 – C \u21d2 c = -37
\n\u2234 Required circle with c = (-2-3) & r = \\(\\sqrt{50}\\) is (x + 2)2<\/sup> + (y – 3)2<\/sup> = \\((\\sqrt{50})^{2}\\)
\nx2<\/sup> + y2<\/sup> + 4x – 6y – 37 = 0.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nFind the equation of the circle passing through the centre of the circle x2<\/sup> + y2<\/sup> – 2x – 4y – 20 = 0 and centre at (4, -2).
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> – 2x – 4y – 20 = 0 & c = (4, -2)
\n\u2234 P= (1, 2)
\nr = \\(\\sqrt{g^{2}+f^{2}-c}=\\sqrt{16+4-(-c)}=\\sqrt{20-c}\\)
\nBut r = CP = \\(\\sqrt{(4-1)^{2}+(-2-2)^{2}}=\\sqrt{3^{2}+(-4)^{2}}=\\sqrt{25}=5\\)
\n\u2234 r = \\(\\sqrt{20-c}\\) \u21d2 5 = \\(\\sqrt{20-c}\\)
\n\u21d2 25 = 20 – c \u21d2 c = -5
\n\u2234 The required equation of the circle is x2<\/sup> + y2<\/sup> – 8x + 4y – 5 = 0.<\/p>\n

Question 5.
\nFind the equation of the circle two of whose diameters are x + y = 3 and 2x + y = 2 and passing through the centre of the circle x2<\/sup> + y2<\/sup> – 4x + 2y – 1 = 0.
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> – 4x + 2y – 1 = 0
\nP = (2,-1)
\n& centre of the circle is point of intersection of diameters
\nx + y = 3 – (1) & 2x + y = 2 – (2)
\neqn. (2) – eqn. (1) gives x = -1 & y = 3 – x = (3 – C – 1) = 4
\n\u2234 centre = c(-1,4)
\nr = CP = \\(\\sqrt{(-1-2)^{2}-\\left(4-(-1)^{2}\\right.}-\\sqrt{(-3)^{2}+5^{2}}=\\sqrt{9+25}=\\sqrt{34}\\)
\n\u2234 the eqn. of the circle with c = (-1,4) & r = \\(\\sqrt{34}\\) is
\n(x + 1)2<\/sup> + (y – 4)2<\/sup> = \\((\\sqrt{34})^{2}\\)
\nx2<\/sup> + 1 + 2x + y2<\/sup> + 16 – 8y = 34
\nx2<\/sup> + y2<\/sup> + 2x – 8y – 17 = 0.<\/p>\n

Question 6.
\nFind the equation of the circle concentric with the centre of circle x2<\/sup> + y2<\/sup> – 2x + 2y – 1 = 0 and having double its area.
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> – 2x + 2y – 1 = 0
\nc = (1, -1) & r\u0131<\/sub> = \\(\\sqrt{1+1-(-1)}=\\sqrt{3}\\)
\nGiven A2<\/sub>, = 2A,
\n\"2nd
\nr2<\/sub>2<\/sup> =2 . \\((\\sqrt{3})^{2}\\) = 2.3 = 6 \u21d2 r2<\/sub> = \\(\\sqrt{6}\\)
\nEquation of the required circle with c = (1,-1) & r = is (x – 1)2<\/sup> + (y + 1)2<\/sup> = \\((\\sqrt{6})^{2}\\).
\nx2<\/sup> + 1 – 2x + y2<\/sup> + 1 + 2y = 6 \u21d2 x2<\/sup> + y2<\/sup> – 2x + 2y – 4 = 0.<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nFind the equation of the circle concentric with the centre of the circle 3x2<\/sup> + 3y2<\/sup> – 6x + 9y – 2 = 0 and \\(\\frac { 2 }{ 3 }\\) having of its area.
\nAnswer:
\nGiven 3x2<\/sup> + 3y2<\/sup> – 6x + 9y – 2 = 0, divide by 3
\nx2<\/sup> + y2<\/sup> – 2x + 3y – \\(\\frac { 2 }{ 3 }\\) =0
\ncentre = (1, \\(\\frac{-3}{2}\\)) & r = \\(\\sqrt{1+\\frac{9}{4}-\\frac{2}{3}}=\\sqrt{\\frac{47}{12}}\\)
\nAlso given A2<\/sub> = \\(\\frac{2}{3}\\) A1<\/sub>
\n\u03c0r2<\/sub>2<\/sup> = \\(\\frac{2}{3}\\) . \u03c0r2<\/sub>2<\/sup> \u21d2 r2<\/sub>2<\/sup> = \\(\\frac{2}{3} \\times \\frac{47}{12}=\\frac{47}{18}\\)
\n\u2234 Required circle with centre \\(\\left(1,-\\frac{3}{2}\\right), r_{2}-\\sqrt{\\frac{47}{18}} \\text { is }(x-1)^{2}+\\left(y+\\frac{3}{2}\\right)^{2}=(\\sqrt{\\frac{47}{18}})^{2}\\)
\nx2<\/sup> + 1 – 2x + y2<\/sup> + 3y + \\(\\frac{9}{4}=\\frac{47}{18}\\)
\nx2<\/sup> + y2<\/sup> – 2x + 3y + \\(\\frac{23}{36}\\) = 0 __________ \u00d7 36
\n36x2<\/sup> + 36y2<\/sup> \u2013 72x + 108y + 23 = 0.<\/p>\n

Question 8.
\nFind the equation of the diameter of the circle x2<\/sup> + y2<\/sup> + 6x – 2y = 6 which when produced passes through the point (1, -2).
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> + 6x – 2y – 6 = 0 & P (1, -2) centre = (-3, 1)
\nSlope of the diameter = m = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{1-(-2)}{-3-1}=\\frac{3}{-4}\\)
\n\u2234 Equation of the diameter with (1, -2) & m = \\(\\frac{-3}{4}\\)
\ny – y1<\/sub> = m(x – x1<\/sub>)
\ny + 2 = \\(\\frac{-3}{4}\\)(x-1)
\n4y + 8 = -3x + 3 \u21d2 3x + 4y + 5 = 0.<\/p>\n

Question 9.
\nFind the equation of the diameter of the circle 2x2<\/sup> + 2y2<\/sup> + 3x – 5y – 1 = 0 which when produced passes through the point (-1,2).
\nAnswer:
\nGiven 2x2<\/sup> + 2y2<\/sup> + 3x – 5y – 1 = 0 & p = (-1,2)
\n\u21d2 x2<\/sup> + y2<\/sup> + \\(\\frac { 3 }{ 2 }\\)x – \\(\\frac { 5 }{ 2 }\\)y – \\(\\frac { 1 }{ 2 }\\) = 0
\nc = \\(\\left(-\\frac{3}{4}, \\frac{5}{4}\\right)\\)
\n\"2nd
\nEquation of the diameter with p(-1,2), & m = -3.
\ny – y1<\/sub>, = m(x – x1<\/sub>); y – 2 = -3(x + 1); y – 2 = -3x \u2013 3; 3x + y + 1 = 0.<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nShow that the line 4x – y=17 passes through the centre of the circle x2<\/sup> + y2<\/sup> – 8x + 2y=0.
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> – 8x + 2y = 0 & line 4x – y = 17 c = (4, -1)
\nC must lie on the line 4x – y = 17
\n4(4)-(-1) = 17
\n16 + 1 = 17 \u21d2 17 = 17
\nHence the line 6x – y = 17 passes through the centre.<\/p>\n

Question 11.
\nFind the length of the chord of the circle x2<\/sup> + y2<\/sup> – 6x + 15y – 16 = 0 intercepted by the x-axis.
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> – 6x + 15y – 6 = 0
\nC = \\(\\left(3, \\frac{-15}{2}\\right)\\) & c = -16
\nLength intercepted by x – axis = 2\\(\\sqrt{g^{2}-c}\\)
\n\\(=2 \\sqrt{(3)^{2}-(-16)}=2 \\sqrt{9+16}=2 \\sqrt{25}=10\\)
\n\u2234 Length = 10 units<\/p>\n

Question 12.
\nFind the length of the chord of the circle x2 <\/sup>+ y2<\/sup> + 3x – y – 6 = 0 intercepted by the y-axis.
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> + 3x – y – 6 = 0
\nc = \\(\\left(-\\frac{3}{2}, \\frac{1}{2}\\right)\\) C=-6
\nLength intercepted by y-axis = 2\\(\\sqrt{f^{2}-c}\\)
\n\"2nd
\n\u2234 Length = 10 units<\/p>\n

Question 13.
\nFind the length of the chord of the circle x2<\/sup> + y2<\/sup> – 6x \u2013 4y – 12 = 0 on the coordinate axes.
\nAnswer:
\nGiven x2<\/sup> + y2<\/sup> – 6x – 4y – 12 = 0
\nC = (3, 2) C = -12
\nLength intercepted by x-axis \\(2 \\sqrt{g^{2}-c}=2 \\sqrt{9+12}=2 \\sqrt{21}\\) units
\nLength intercepted by y-axis = \\(2 \\sqrt{f^{2}-c}=2 \\sqrt{4+12}=2 \\sqrt{16}\\) = 8units.<\/p>\n

\"KSEEB<\/p>\n

Part – C<\/span><\/p>\n

2nd PUC Basic Maths Circles Ex 15.2 Five Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nFind the equation of the circle.
\n(a) Passing through the origin, having its centre on the x- axis and radius 2 units.
\n(b) Passing through (2, 3) having its centre on the x-axis and radius 5 units.
\n(c) Passing through the points (5, 1), (3, 4) and has its centre on the x – axis.
\n(d) Passing through the points (1, 2) and (2, 1) and has its centre on the y – axis.
\n(e) Passing through the points (0,5) and (6, 1) and has its centre on the line 12x + 5y = 25.
\n(f) Passing through the points (1, -4) and (5,2) and has its centre on the line x – 2y + 9 =0.
\n(g) Passing through the points (0, -3) and (0,5) and whose centre lies on x – 2y + 5 = 0.
\n(h) Passing through (1, 1) and (2, 2) and having radius 1.
\nAnswer:
\n(a) Let A = (0,0), centre on x – axis \u21d2 C = (-g, 0), r = 2 units
\nRequired equation is x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\nIt passes their (0,0) \u21d2 c = 0, r = \\(\\sqrt{g^{2}+f^{2}-c}\\)
\n2 = \\(\\sqrt{g^{2}+0-0}\\) \u21d2 g = \u00b12
\nEquation of the circle is x2<\/sup> + y2<\/sup> \u00b1 4x = 0.<\/p>\n

(b) Let the general equation of the circle be x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0.
\nIt passes through (2,3) \u21d2 (2)2<\/sup> + (3)2<\/sup> + 2g(2) + 2f(3) + c = 0
\n4g + 6f + c + 13 = 0 ….(1)
\nCentre (-g, -f) lies on x – axis \u21d2 f = 0 …(2)
\nr = \\(\\sqrt{g^{2}+f^{2}-c}\\) = 5 = \\(\\sqrt{g^{2}-c}\\) = 25 = g2<\/sup> – C.
\n\u21d2 g2<\/sup> \u2013 25 = c. ….(3)
\nSolving 1, 2 & 3 we get g2<\/sup> – 25 + 4g + 13 = 0
\ng2<\/sup> + 4g – 12 = 0
\n\u21d2 (g + 6) (g – 2) = 0 \u21d2 g = -6 or 2
\nWhen g = -6, c = (-6)2<\/sup> – 25 = 36 – 25 = 11
\ng = 2, c = (2)2<\/sup> – 25 = 4 – 25 = -21 .
\n\u2234 The required circles are x2<\/sup> + y2<\/sup> – 12x + 11 = 0
\nx2<\/sup> + y2<\/sup> + 4x – 21 = 0.<\/p>\n

\"KSEEB<\/p>\n

(C) Let the required equation of the circle is x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\nIt passes through (5, 1) & (3, 4)
\n(5, 1) (5)2<\/sup> + (1)2<\/sup> + 2g(5) + 2f (1) +C = 0
\n10g + 2f + C + 26 = 0 …..(1)
\n(3,4) 32<\/sup> + 42<\/sup> + 2g (3) + 2f (4) +c= 0
\n68 + 8f + C + 25 = 0 …..(2)
\n& the centre (- g, – f) lies on x-axis \u21d2 f = 0 ….(3)
\nSolving 1 & 2 we get
\n\"2nd
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – x – 47 = 0.<\/p>\n

(d) Let required equation of the circle is x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\nIt passes through the points (1,2) & (2, 1) & centre lies on y-axis \u21d2 g = 0 ….(1)
\n(1,2) 12<\/sup> + 22<\/sup> + 2g(1) + 2f(2) + c = 0 \u21d2 4f + C + 5 = 0 ….(2)
\n(2, 1) 22<\/sup> + 12<\/sup> + 2g(2) + 2f(1) + c = 0 \u21d2 2f+c + 5 = 0 ….(3)
\nEquation (3) – eqn. (2) gives 2f = 0 \u21d2 f = 0
\ng = 0, f = 0 \u21d2 c= -5
\n\u2234 Required equation of the circle is x2<\/sup> + y2<\/sup> – 5 = 0<\/p>\n

(e) Let required equation of the circle is x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\nIt passes through (0,5) & (6, 1)
\n(0,5) 0 + 52<\/sup> + 2g(0) + 2f(5) + c = 0
\n1f + C + 25 = 0 …(1)
\n(5, 1) 62<\/sup> + 12<\/sup> + 2g (6) + 2f (1) + c = 0
\n12g + 2f + c + 37 = 0 …..(2)
\nCentre (-g, -f) lies on the line 12x + 5y – 25 = 0
\n-12g – 5f – 25 = 0 ….(3)
\nEqn. 2 – eqn. (1) gives
\n12g – 8f + 12 = 0 …..(4)
\nAdding 3 & 4 we get
\n12g – 13f – 13 = 0 \u21d2 f = -1
\nFrom eqn<\/sup> (1) \u21d2 10(-1) + C + 25 = 0 \u21d2 C = – 15
\n\u2234 -12g = 5f + 25
\n-12g = 20 \u21d2 g = \\(\\frac{20}{-12}=\\frac{-5}{3}\\)
\n\u2234 The required equation of the circle is
\nx2<\/sup> + y2<\/sup> + 2\\(\\left(-\\frac{5}{3}\\right)\\)x + 2(-1)y \u2013 15 = 0 \u21d2 \u00d7 3
\n3x2<\/sup> + 3y2<\/sup> – 10x – 6y – 45 = 0<\/p>\n

\"KSEEB<\/p>\n

(f) Let the required equation of the circle is x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 but it passes through (1,-4) & (5,2) & centre (-g, -f) lies on the line
\n\u21d2 -g + 2f + 9 = 0 …… (1)
\n(1,-4) 12<\/sup> + (-4)2<\/sup> + 2g(1) + 2f(-4) + c = 0
\n2g – 8f + C + 17 = 0 ….. (2)
\n\u21d2 (5,2) 52<\/sup> + 22<\/sup> + 2g(5) + 2f(2) +C = 0
\n10g + 4f + C+ 29 = 0 ….. (3)
\nEqn. (3) – eqn. (2) gives 8g + 12f + 12 = () – 4
\n2g + 3f + 3 = 0 …..(4)
\n\"2nd
\n2f + 9 = g \u21d2 g = -6 + 9 = 3
\nC = -2g + 8f – 17
\nC = -6 -24 – 17 = – 47
\n\u2234 The required equation of the circle is
\nx2<\/sup> + y2<\/sup> + 2 (3)x + 2 (-3)y – 47 = 0
\nx2<\/sup> + y2<\/sup> + 6x – 6y – 47 = 0<\/p>\n

(g) Let the equation of required circle is x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 it passes
\nThrough (0, -3) & (0,5) & centre (-g, -f)
\nLies on the line x – 2y + 5 = 0 \u21d2 -g + 2f + 5 = 0 …..(1)
\n(0,-3) 0 + 9 + (-6f) + c = 0
\n-6f + 9 + c = 0 …..(2)
\n(0,5) 0 + 25 + 0 + 10f + C = 0
\n10f + C + 25 = 0 …. (3)
\n\u2234 eqn. (3) – eqn. (2) gives 16f + 16 = 0 \u21d2 f =-1
\n2f + 5 = g \u21d2 g= -2 + 5 = 3
\nc = 6f – 9 = -6 -9 = -15 \u21d2 c = -15
\n\u2234 The equation of the circle with g = 3, f = -1 & C = -15 is
\nx2<\/sup> + y2<\/sup> + 2 (3) x + 2 (-1)y – 15 = 0
\nx2<\/sup> + y2<\/sup> + 6x – 2y – 15 = 0<\/p>\n

(h) Let the equation of the required circle is x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\nIf passes through (1, 1) & (2,2)
\n(1,1) 12<\/sup> + 12<\/sup> + 2g(1) + 2f(1) + c = 0
\n2g + 2f + C + 2 = 0 ….. (1)
\n(2, 2) 4 + 4 + 2g(2) + 2f (2) + c = 0
\n4g + 4f + c + 8 = 0 …… (2)
\nAlso r = 1 \u21d2 \\(\\sqrt{g^{2}+f^{2}-c}\\) =1 \u21d2 g2<\/sup> + f2<\/sup> – c = 1 …… (3)
\nEqn. (2) – eqn. (1) gives 2g + 2f + 6 = 0
\ng + f + 3 = 0
\ng = (-3 -f) ….(4)
\nAdding 1 and 3 we get
\ng2<\/sup> + f2<\/sup> + 2g + 2f = -1
\n(-3 -f)2<\/sup> + f2<\/sup> + 2 (-3-f) + 2f + 1 = 0
\n\"2nd
\n2f2<\/sup> + 6f + 4 = 0 \u00f7 2
\nf2<\/sup> + 3f + 2 = 0
\n(f + 2) (f + 1) = 0 \u21d2 f = -1 or -2
\nWhen f = -2, g = -3-f = -3 + 2 = -1
\nWhen f = -1, g = -3-f = -3 + 1 = -2
\nC = g2<\/sup>+ f2<\/sup> -1 = 1 + 4 – 1 = 4
\n\u2234 Required circles are x2<\/sup> + y2<\/sup> – 2x – 4y + 4 = 0 and x2<\/sup> + y2<\/sup> – 4x – 2y + 4 = 0.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind the equation of the circle passing through the points.
\n(a) (0, 2), (3, 0), (3, 2)
\n(b) (1, 1), (-2, 2), (-6,0)
\n(c) (1, 1), (5,-5), (6,-4)
\n(d) (1, 0), (3,0), (0, 2)
\n(e) (5, 7), (6, 6), (2,-2)
\n(0) (p, q), (p, 0), (0,)
\n(g) (0, 1), (2,3), (-2,5)
\n(h) (0,0), (a,0), (0, b)
\nAnswer:
\n(a) Let the required equation of the circle is x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\nThis equation passes through (*0,2) (3,0) & (3,2)
\n(0,2) 02<\/sup> + 22<\/sup> + 2g(0) + 2 f(2) + c = 0
\n4f + 4 + c = 0 …. (1)
\n(3,0) 6g + 9 + c = 0 …. (2)
\n(3,2) 6g + 4f + 13 + c = 0 …. (3)
\nSolving (1), (2), (3) we get
\n(2) – (1) gives g – 4f + 5 = 0
\n(3) – (2) gives 4f + 4 = 0 f = -1
\n6g = 4(-1) -5 = -9 \u21d2 g =\\(-\\frac{9}{6}=-\\frac{3}{2}\\).
\nC = -4f – 4 = 4 – 4 = 0
\n\u2234 Required equation of the circle is
\n\"2nd<\/p>\n

(b) (1, 1) (-2,2) (-6,0)
\nThe general equation of the circle x2<\/sup> + y2<\/sup> + 2gx + 2fy +C = 0
\nThis equation passes through (1, 1) (-2,2) & (-6,0)
\n(1, 1) 1 + 1 + 2g + 2f + c = 0
\n2g + 2f + 2 + c = 0 …. (1)
\n(-2,2) 4 + 4 – 4g + 4f + c = 0
\n-4g + 4f + 8 + c ….. (2)
\n(-6,0) 36 + 0 – 12g + c = 0
\n-12g + 36 + c = 0 …. (3)
\n(2) – (1) gives -6g + 2f + 6 = 0 … (4) \u00d7 2
\n\"2nd
\n\u2234 The required eqn. is x2<\/sup> + y2<\/sup> + 4x + 6y – 12 = 0<\/p>\n

\"KSEEB<\/p>\n

(c) (1, 1) (5,-5) (6,-4)
\nThe general equation of the circle
\nx2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\nThis equation passes through (1, 1) (5,-5) (6, -4)
\n(1; 1) 2g + 2f + C + 2 = 0 ….. (1)
\n(5,-5) 10g – 10f + C + 50 = 0 …. (2)
\n(6,4) 12g + 8f + C + 52 = 0 …. (3)
\n(2)-(1) gives 8g – 12f + 48 = 0 ….. (4)
\n(3) – (2) gives 2g + 18f + 2 = 0 …. (5)
\nSolving (4) & (5) we get f = \\(\\frac { 10 }{ 11 }\\) g = \\(\\frac { 111 }{ 21 }\\) & C = \\(-\\frac { 284 }{ 21 }\\)
\n\u2234 the required equation is
\nx2<\/sup> + y2<\/sup> + \\(\\frac { 222x }{ 21 }\\) + \\(\\frac { 20y }{ 21 }\\) – \\(\\frac { 284 }{ 21 }\\) = 0 – \u00d7 21
\n21x2<\/sup> + 21y2<\/sup> + 222x + 20y – 284 = 0<\/p>\n

(d) Let the general eqn, x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0. The circle passes through the points
\n(1, 0) (3,0) & (0,2)
\n(1,0) 2g + C + 1 = 0 …. (1)
\n(0,2) 4f + C + 4 = 0 ….. (2)
\n(3,0) 6g + C + 9 = 0 …… (3)
\nSolving 1, 2, 3 we get g = -2, f = \\(-\\frac{7}{4}\\) & c = 3
\n\"2nd
\n\u2234 2x2<\/sup> + 2y2<\/sup> – 8x \u2013 7y + 6 = 0.<\/p>\n

\"KSEEB<\/p>\n

(e) Let the general eqn. x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0. The circle passes through the points
\n(5,7) (6, 6) (2,-2)
\n(5,7) 10g + 14f + 74 + c = 0 … (1)
\n(6,6) 12g + 12f + C + 72 = 0 …(2)
\n(2,-2) 4g – 4f + C + 8 = 0 …(3)
\nSolving 1, 2 and 3 we get g = -2, f = -3 & c= -12
\n\u2234 Required equation is x2<\/sup> + y2<\/sup> – 4x – 6y – 12 = 0.<\/p>\n

(f) (p, q) (p, 0) (0,9)
\nLet the general eqn. x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0.
\nThe circle passes through the points
\n(p, q) (p, 0) (0,q)
\n(p, q) p2<\/sup> + q2<\/sup> + 2pq + 2qf + c = 0 … (1)
\n(p, 0) p2<\/sup> + 2pg + c = 0 …… (2)
\n(0,q) q2<\/sup> + 2qf + c= 0 … (3)
\n1 – 2 gives 2qf = -q2<\/sup> \u21d2 f = \\(-\\frac{q}{2}\\)
\n1 – 3 gives 2pg = -p2<\/sup> \u21d2 g= \\(-\\frac{p}{2}\\) & c = 0
\n\u2234 Required eqn. is x2<\/sup> + y2<\/sup> – px – qy = 0.<\/p>\n

(g) Let the general eqn, x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0. The circle passes through the points.
\n(0, 1) (2,3) (-2,5)
\n(0, 1) 2f + C = -1 ….. (1)
\n(2,3) 4g + 6 + c = -13 … (2)
\n(-2,5) -4g + 10f +c= -29 … (3)
\nSolving 1, 2, 3 we get g = \\(\\frac{1}{3}\\) , f = \\(-\\frac{10}{3}\\), c = \\(\\frac{17}{3}\\)
\n\u2234 Required equation is x2<\/sup> + y2<\/sup> + \\(\\frac{2}{3}\\)x – \\(\\frac{20}{3}\\)y + \\(\\frac{17}{3}\\) = 0
\n3x2<\/sup> + y2<\/sup> + 2x – 20y + 17 = 0<\/p>\n

\"KSEEB<\/p>\n

(h) Let the general equation x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0. The circle passes through the points (0,0) (a, 0) (0, b)
\n(0,0) c = 0 …. (1)
\n(a,0) 2ag + c = -a2<\/sup> \u21d2 g = \\(-\\frac{a}{2}\\)
\n(0, b) 2bf + c = -b2<\/sup> \u21d2 f = \\(-\\frac{b}{2}\\)
\n\u2234 Required equation is x2<\/sup> + y2<\/sup> – ax – by = 0.<\/p>\n

Question 3.
\nFind the equations of the circles whose radius is 5 and which passes through the points on X-axis at distances 3 from the origin.
\nAnswer:
\nGiven r = 5, passes through points on x-axis at a distance 3 from the origin
\n\u21d2 The points are (3,0) and (-3,0)
\nLet x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 is required equation
\n(3,0) 6g + C + 9 = 0 … (1)
\n(-3,0) -6g + C + 9 = 0 …. (2)
\nr = 5; 25 = g2<\/sup> + f2<\/sup> -C
\nAdding 1 and 2 we get c = -9 and g = 0 and f = \u00b14
\n\u2234 The required equation is x2<\/sup> + y2<\/sup> + 8y – 9 = 0<\/p>\n

Question 4.
\nA circle has radius 3 units and its centre lies on the line y = x – 1. Find the equation of the circle if it passes through (7,3).
\nAnswer:
\nGiven r = 3 and centre (-g,-f) lies on y = x – 1
\n\u21d2 -f = -g -1
\ng – f + 1 = 0 …. (1)
\nEquation passes through (7,3) \u21d2 14g + 6f + C + 58 = 0 …. (2)
\nr = 3 \u21d2 r2<\/sup> = g2<\/sup> + f2<\/sup> -c = 9 … (3)
\nSolving 1, 2, 3 we get g = -7 or g = -4, c = 76
\nf = -6 or g = -3, c = 16
\n\u2234 We get two equations of circles
\nx2<\/sup> + y2<\/sup> -14x \u2013 12y + 76 = 0
\nx2<\/sup> + y2<\/sup> – 8x – 6y + 16 = 0.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nFind the equation of the circle cuts intercepts of the length ‘a’ and ‘b’ on axes and passes through the origin
\nAnswer:
\nIntercepts a and b on axes that implies circle passes through the points (a, 0) and (0, b) passing through origin
\n\u21d2 c = 0
\n(a, 0) a2<\/sup> + 2ga = 0 \u21d2 g = \\(-\\frac{a}{2}\\)
\n(0, -b) b2<\/sup> + 2fb = 0 \u21d2 f = \\(-\\frac{b}{2}\\)
\n\u2234 The required equation is x2<\/sup> + y2<\/sup> – ax – by = 0.<\/p>\n

Question 6.
\nFind the equation of the circle passing through the origin and lengths ‘a’ and ‘b’ on axes and passing through the origin.
\nAnswer:
\nPositive x-intercept a i.e., (a,0)
\nNegative y-intercept bi.e., (0, -b)
\nLet x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 passes through origin \u21d2 c = 0 …. (1)
\n(a,0) \u21d2 a2<\/sup> + 2ga = 0 = g \u21d2 \\(-\\frac{a}{2}\\) .. (2)
\n(0, -b) \u21d2 b2<\/sup> – 2fb = 0 = f \u21d2 \\(-\\frac{b}{2}\\) … (3)
\n\u2234 The required equation is x2<\/sup> + y2<\/sup> – ax + by = 0.<\/p>\n

Question 7.
\nFind the length of the chord intercepted by the
\n(a) circle x2<\/sup> + y2<\/sup> – 8x – 6y = 0 and the line x – 7y – 8 = 0
\n(b) circle x2<\/sup> + y2<\/sup> = 9 and the line x + 2y = 3
\n(c) circle x2<\/sup> + y2<\/sup> – 6x – 2y + 5 = 0 and the line x – y+ 1 = 0
\nAnswer:
\n(a) Given centre (4,3)
\nr= \\(=\\sqrt{16+9}=\\sqrt{25}\\) = 5
\nP = length of the perpendicular from (4, 3) to the line x – 7y – 8 = 0
\n\"2nd
\nLength of the chord = 2 . \\(\\sqrt{r^{2}-p^{2}}\\)
\n\\(=2 \\sqrt{5^{2}-\\left(\\frac{5}{\\sqrt{2}}\\right)^{2}}=2 \\sqrt{25-\\frac{25}{2}}=2 \\sqrt{\\frac{50-25}{2}}=2 \\cdot \\frac{5}{\\sqrt{2}}\\)
\nL = 5\\(\\sqrt{2}\\) units<\/p>\n

\"KSEEB<\/p>\n

(b) Centre (0,0), r = 3, Line x + 2y – 3 = 0
\n\"2nd
\nLength of the chord = \\(2 \\cdot \\sqrt{r^{2} \\cdot d^{2}}=2 \\sqrt{9-\\frac{9}{5}}=2 \\sqrt{\\frac{36}{5}}=\\frac{12}{\\sqrt{5}}\\) units<\/p>\n

(c) Centre = (3, 1) and r = \\(\\sqrt{9+1-5}=\\sqrt{5}\\) , line x – y + 1 = 0
\n\"2nd
\nLength of the chord = \\(2 \\sqrt{r^{2}-d^{2}}=2 \\sqrt{5-\\frac{9}{2}}=2 \\cdot \\frac{1}{\\sqrt{2}} 5 \\text { units }\\)<\/p>\n

Question 8.
\nFind the points of intersection of the circle.
\n(a) x2<\/sup> + y2<\/sup> = 9 and the line x + 2y = 3.
\n(b) x2<\/sup> + y2<\/sup> – 6x – 2y + 5 = 0 and the line x – y + 1=0.
\n(c) x2<\/sup> + y2<\/sup> + 4x + 6y – 12 = 0 and the line x + 4y – 6 = 0
\nAlso find the length of the chord.
\nAnswer:
\n(a) Given x = 3 – 2y and x2<\/sup> + y2<\/sup> = 9
\n(3 – 2y)2<\/sup> + y2<\/sup> = 9
\n5y2<\/sup> – 12y = 0
\ny(5y – 12) = 0 \u21d2 y = 0, y = 12
\n\u21d2 x = 3, x = \\(-\\frac{9}{5}\\)
\nThe points are (3, 0) and \\(\\left(-\\frac{9}{5}, \\frac{12}{5}\\right)\\)
\n\u2234 Length of the chord = \\(\\sqrt{\\left(3+\\frac{9}{5}\\right)^{2}+\\left(\\frac{12}{5}\\right)^{2}}=\\frac{\\sqrt{720}}{5}\\)<\/p>\n

(b) Given
\nx + 1 = y and x2<\/sup> + y2<\/sup> – 6x – 2y + 5 = 0
\nx2<\/sup> + (x + 1)2<\/sup> – 6x – 2 (x + 1) + 5 = 0
\n2x2<\/sup> – 6x + 4 = 0
\nx2<\/sup> – 3x + 2 = 0
\n(x – 2)(x – 1) = 0 = x= 2 or 1
\nWhen x = 2, y = 3 and when x = 1, y = 2
\n\u2234 The points are (2, 3) and (1, 2) and length of the chord
\n= \\(\\sqrt{(2-1)^{2}+(3-2)^{2}}=\\sqrt{1+1}=\\sqrt{2}\\)<\/p>\n

\"KSEEB<\/p>\n

(c) Given x = 6 – 4y and x2<\/sup> + y2<\/sup> + 4x + 6y – 12 = 0
\n(6 – 4y)2<\/sup> + y2<\/sup> + 4 (6 – 4y) + 6y – 12 = 0
\n17y2<\/sup> – 58y + 48 = 0
\n17y2<\/sup> – 34y – 24y + 48 = 0
\n17y (y – 2) – 24 (y – 2) = 0
\n(y – 2) (17 y – 24) = 0 \u21d2 y = 2 or
\nWhen y = 2, x = 6 – 4y = 6 – 8 = -2
\nWhen y = \\(\\frac { 24 }{ 17 }\\) x = \\(6-\\frac{96}{17}=\\frac{102-96}{17}=\\frac{6}{17}\\)
\n\u2234 The points are (-2,2) and \\(\\left(\\frac{6}{17}, \\frac{24}{17}\\right)\\)
\n\"2nd<\/p>\n

Part-D<\/span><\/p>\n

2nd PUC Basic Maths Circles Ex 15.2 Six Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nShow that the following points are concylie.
\n(a) (0, 0), (1, 1), (5, -5), (6,-4)
\n(b) (2,-4), (3, -1), (3,-3), (0,0)
\n(c) (1.0), (2, -7), (8, 1), (9,-6)
\nAnswer:
\n(a) First find the equation of the circle passing through the points (0,0) (1, 1) (5,-5) (6, -4)
\n(0,0) \u21d2 C = 0 …. (1)
\n(1, 1) \u21d2 2g + 2f + 2 = 0 …. (2)
\n(5,-5) \u21d2 10g -10f + 50 + c = 0 …. (3)
\n\"2nd
\n\u2234 The equation of the circle is x2<\/sup> + y2<\/sup> – 6x + 4y = 0
\nSubstituting the fourth point (6,-4) we get
\n36 + 16 – 36 – 16 = 0 \u21d2 0 = 0
\n\u2234 The points are concyclic.<\/p>\n

(b) Let us find the equation of the circle passing through (2,-4) (3,-1) and (3, -3) we get
\n(2,-4) 4g – 8f + C + 20 = 0 … (1)
\n(3,-1) 6g – 2f+ C + 10 = 0) … (2)
\n(3,-3) 6g – 6f+c= -18 … (3)
\nSolving the above equations we get g = -1, f = 2 and C = 0
\n\u2234 Required circle is x2<\/sup> + y2<\/sup> – 2x + 4y = 0
\nSubstitute the fourth point (0,0), we get 0 + 0 + 0 + 0 = 0
\n\u2234 The four points are concyclic<\/p>\n

\"KSEEB<\/p>\n

(c) Let us find the equation of the circle passing through (1, 0) (2, -7) and (8, 1) we get
\n(1,0) 2g + C + 2 = 0 … (1)
\n(2,-7) 4g – 14f + C + 53 = 0 … (2)
\n(8,1) 16g + 2F + C + 25 = 0 … (3)
\nSolving above 3 equations we get g = \\(-\\frac { 738 }{ 50 }\\),f = \\(\\frac { 147 }{ 50 }\\) and c = \\(\\frac { 196 }{ 25 }\\)
\n\u2234 Required circle is x2<\/sup> + y2<\/sup> + \\(\\frac{246}{50} x\\) + \\(\\frac{147}{25} y\\) + \\(\\frac{196}{25} \\) = 0
\n25x2<\/sup> + 25y2<\/sup> + 246 x + 147 y + 196 = 0
\nSubstitute the fourth point (9,-6) we get
\n25(9)2<\/sup> + 25(-6)2<\/sup> + 246 (9) + 147 (-6) + 196 = 0
\n2025 + 900 + 2214 – 882 + 196 0
\nThe points are not concyclic<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 15.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.2 Part – A 2nd PUC Basic …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20351"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=20351"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20351\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20351"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20351"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20351"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}