{"id":20324,"date":"2020-01-20T15:54:03","date_gmt":"2020-01-20T10:24:03","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20324"},"modified":"2021-07-02T15:28:16","modified_gmt":"2021-07-02T09:58:16","slug":"2nd-puc-basic-maths-question-bank-chapter-15-ex-15-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-15-ex-15-1\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.1"},"content":{"rendered":"

Students can Download Basic Maths Exercise 15.1 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 15.1<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Circles Ex 15.1 One Mark Questions and Answers<\/strong><\/p>\n

Question 1.
\nFind the equation of the circle with centre ‘C’ and radius ‘r’ in each of the following:
\n(a) c(-3, 2) and r= 5 units
\n(b) c(0, 0) and r = 4 units
\n(c) c(-1,-2) and diameter = 25 units
\n(d) c(1, 1) and r = \\(\\sqrt{2}\\) units
\n(e) c(a cos \u03b8, b sin \u03b8) and r = a units.
\nAnswer:
\n(a)
\n\"2nd
\nEquation of the circle is(x – x1<\/sub>)2<\/sup> + (y – y1<\/sub>)2<\/sup> = r2<\/sup>
\n(x-(-3))2<\/sup> + (y – 2)2<\/sup> = 52<\/sup>
\n(x + 3)2<\/sup> + (y – 2)2<\/sup> = 52<\/sup>
\nx2<\/sup> + y2<\/sup> + 6x – 4y – 12 = 0.<\/p>\n

(b) Given C = (0,0) & r = 4 units
\nThe equation of the circle is
\n(x – 0)2<\/sup> + (y – 0)2<\/sup> = 42<\/sup>
\n\u21d2\u00a0x2<\/sup> + y2<\/sup> = 16<\/p>\n

\"KSEEB<\/p>\n

(c) C = (-1,-2 )& r = \\(\\frac{\\text { diameter }}{2}=\\frac{25}{2}\\)
\nThe equation of the circle is
\n(x + 1)2<\/sup> + (y + 2)2<\/sup> = \\(\\left(\\frac{25}{2}\\right)^{2}\\)
\nx2<\/sup> + 2x + 1 + y2<\/sup> + 4y +4 = \\(\\frac { 625 }{ 4 }\\)
\n\u21d2 4x2<\/sup> + 4y2<\/sup> + 8x + 16y – 605 = 0.<\/p>\n

(d) C = (1, 1) r = \\(\\sqrt{2}\\) units
\nEquation of the circle is
\n(x – 1)2<\/sup> +(y – 1)2<\/sup> = \\((\\sqrt{2})^{2}\\)
\n\u21d2 x2<\/sup> + 1 – 2x + y2<\/sup> + 1 – 2y = 2
\n\u21d2 x2<\/sup> + y2<\/sup> – 2x – 2y = 0<\/p>\n

(e) C = (a cos \u03b8, a sin \u03b8) and r = a units
\nEquation of the circle is (x – a cos \u03b8)2<\/sup>+ (y – a sin \u03b8)2<\/sup> = a2<\/sup>
\n\u21d2 x2<\/sup> + a2<\/sup> cos2<\/sup> \u03b8 – 2a x cos \u03b8 + y2<\/sup> + a2<\/sup> sin2<\/sup> \u03b8 – 2a y sin \u03b8 = a2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> – 2a x cos \u03b8 – 2 ay sin \u03b8 + a2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8) = a2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> – 2a x cos \u03b8 – 2a y sin \u03b8 + a2<\/sup> – a2<\/sup>= \u03b8
\n\u21d2 x2<\/sup> + y2<\/sup> – 2a x cos \u03b8 – 2a y sin \u03b8 = 0 (\u2235\u00a0 cos2<\/sup> \u03b8+ sin2<\/sup> \u03b8 = 1)<\/p>\n

Question 2.
\nFind the centre of the circle, two of the diameters are
\n(a) x + y = 2, x – y = 0
\n(b) 2x – 3y = 1, 3x – 2y = 2
\n(c) y = 0 and y = x – 5
\nAnswer:
\n(a) Given x + y = 2, x – y = 0 solving these two equations we get
\n\"2nd<\/p>\n

(b) Given 2x – 3y = 1 \u2192 \u00d7 3
\n3x – 2y = 2 \u2192 \u00d7 2
\n\"2nd<\/p>\n

(c) y = 0 \u21d2 0 = x – 5 \u21d2 x = 5
\n\u2234 centre = (5,0).<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nWrite the equation of the point circle with centre at
\n(a) (4, -5)
\n(b) (-3, 2)
\n(c) (1,0).
\nAnswer:
\n(a) Point circle means circle whose radius = 0
\nGiven C = (4,-5) & r = 0
\nThe point circle & (x – 4)2<\/sup> + (y + 5)2<\/sup> = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 8x + 10y + 41 = 0<\/p>\n

(b) Centre (-3,2)
\nThe point circle is (x + 3)2<\/sup> + (y – 2)2<\/sup> = 0
\n\u21d2 x2<\/sup> + y2<\/sup> + 6x – 4y + 13 = 0<\/p>\n

(C) Centre (1,0)
\n\u2234 Circle is (x – 1)2<\/sup> + (y – 0)2<\/sup> = 0
\nx2<\/sup> – 2x + 1 + y2<\/sup> = 0 \u21d2 x2<\/sup> + y2<\/sup> – 2x + 1 = 0.<\/p>\n

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

2nd PUC Basic Maths Circles Ex 15.1 Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nFind the equation of the circle.
\n(a) Two of the diameters are x + y = 6 and x + 2y = 4 and its radius is 10 units.
\n(b) Two of the diameters x + y = 4 and x – y = 2 and passing through the point (2,-1).
\n(c) Two of the diameters are 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 Sq. cm.
\nAnswer:
\n(a) Given r = 10, & Diameters are x + y = 6 & x + 2y = 4
\nSolving
\n\"2nd
\n\u2234 Centre = (8,-2) & r = 10 units.
\n\u2234 Equation of the circle is (x – 8)2 <\/sup>+ (y + 2)2<\/sup> = 10
\n\u21d2 x2<\/sup> + y2<\/sup> – 16x + 4y – 32 = 0.<\/p>\n

\"KSEEB<\/p>\n

(b) Point of inter section of two diameters x + y = 4 & x – y = 2 is the centre.
\n\"2nd
\n\u2234 Centre C = (3, 1). Also given P(2, – 1)
\nr = CP = \\(\\sqrt{(2-3)^{2}-(-1-1)^{2}}=\\sqrt{(-1)^{2}+(-2)^{2}}=\\sqrt{1+4}=\\sqrt{5}\\)
\n\u2234 The required equation of the circle with centre (3, 1) & radius = \\(\\sqrt{5}\\) is
\n(x – 3)2<\/sup> – (y – 1)2<\/sup> = \\((\\sqrt{5})^{2}\\)
\nx2<\/sup> + 9 – 6x + y2<\/sup> + 1 – 2y = 5
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x – 2y + 5 = 0.<\/p>\n

(c) Point of intersection of the two diameters is centre
\n2x – 3y = 5 \u2192 \u00d7 3
\n3x – 4y = 7 \u2192 \u00d7 2
\n\"2nd
\n\u2234 C = (1, – 1)
\nAlso Given Area of the circle = 154 sq. cm
\n\u21d2 \u03c0r2<\/sup> = 154
\n\\(r^{2}=\\frac{154}{\\pi}=\\frac{154}{22} \\times 7=49\\)
\n\u21d2 r = 7
\n\u2234 Required Equation of the circle with centre (1, -1) & r = 7 units is
\n(x – 1)2<\/sup> + (y + 1)2<\/sup> = 49
\nx2<\/sup> + 1 – 2x + y2<\/sup> + 2y + 1 = 49
\nx2<\/sup> + y2<\/sup> – 2x + 2y – 47 = 0.<\/p>\n

Question 2.
\nFind the equation of the circle with centre at (-2, 1) and passing through the origin.
\nAnswer:
\nGiven C = (-2, 1) & Let P = (0,0)
\nr = CP = \\(\\sqrt{(0-(-2))^{2}+(0-1)^{2}}=\\sqrt{2^{2}+(-1)^{2}}=\\sqrt{4+1}=\\sqrt{5}\\)
\n\u2234 Required equation of the circle with centre C = (-2, 1) & r = \\(\\sqrt{5}\\) units is
\n(x + 2)2<\/sup> + (y – 1)2<\/sup> = \\((\\sqrt{5})^{2}\\)
\n\u21d2 x2<\/sup> + y2<\/sup> + 4x – 2y = 0.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFind the equation of the circle with centre at (2, 1) and passing through (0, -1).
\nAnswer:
\nGiven Centre = C (2, 1) & Let P = (0, -1)
\nHere r = CP = \\(\\sqrt{(0-2)^{2}+(-1-1)^{2}}=\\sqrt{4+4}=\\sqrt{8}\\) units
\nEquation of the circle with centre (2, 1) & r = \\(\\sqrt{8}\\) is
\n(x – 2)2<\/sup> + (y – 1)2<\/sup> = \\((\\sqrt{8})^{2}\\)
\nx2<\/sup> + 4x + y2<\/sup> + 1 – 2y = 8
\nx2<\/sup> + y2<\/sup> – 4x – 2y – 3 = 0.<\/p>\n

Question 4.
\nFind the equation of the circle described on the line joining the points A and B as diameter where.
\n(a) A(-5, 1) and B(1,3)
\n(b) A(2,0) and B(0, 2)
\n(c) A(3,4) and B(1, -2)
\nAnswer:
\n(a)
\n\"2nd
\nEquation of the circle is (x – x1<\/sub>) (x – x2<\/sub>) + (y – y1<\/sub>) (y – y2<\/sub>) = 0
\ni.e., (x + 5) (x – 1) + (y – 1) (y – 3) = 0
\n\u21d2 x2<\/sup> + 5x – x – 5 + y2<\/sup> – 3y – y + 3 = 0
\n\u21d2 x2<\/sup> + y2<\/sup> + 4x – 4y – 2 = 0.<\/p>\n

(b)
\n\"2nd
\nEquation of the circle is (x – x1<\/sub>)(x – x2<\/sub>) + (y – y1<\/sub>) (y – y2<\/sub>) = 0
\ni.e., (x – 2)(x – 0) + (y – 0) (y – 2) = 0
\nx2<\/sup> – 2x + y2<\/sup> – 2y = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 2x – 2y = 0<\/p>\n

(c) Given A(3,4) B(1, -2)
\n\u2234\u00a0Required equation is (x – x1<\/sub>)(x – x2<\/sub>) + (y – y1<\/sub>) (y – y2<\/sub>) = 0
\ni.e., (x – 3) (x – 1) + (y – 4) (y + 2) = 0
\nx2<\/sup> – x – 3x + 3 + y2<\/sup> + 2y – 4y – 8 = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 4x – 2y – 5 = 0.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 15.1 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 15.1 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\/20324"}],"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=20324"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20324\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20324"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20324"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20324"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}