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## Karnataka 2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.1

Part – A

**2nd PUC Basic Maths Circles Ex 15.1 One Mark Questions and Answers**

Question 1.

Find the equation of the circle with centre ‘C’ and radius ‘r’ in each of the following:

(a) c(-3, 2) and r= 5 units

(b) c(0, 0) and r = 4 units

(c) c(-1,-2) and diameter = 25 units

(d) c(1, 1) and r = \(\sqrt{2}\) units

(e) c(a cos θ, b sin θ) and r = a units.

Answer:

(a)

Equation of the circle is(x – x_{1})^{2} + (y – y_{1})^{2} = r^{2}

(x-(-3))^{2} + (y – 2)^{2} = 5^{2}

(x + 3)^{2} + (y – 2)^{2} = 5^{2}

x^{2} + y^{2} + 6x – 4y – 12 = 0.

(b) Given C = (0,0) & r = 4 units

The equation of the circle is

(x – 0)^{2} + (y – 0)^{2} = 4^{2}

⇒ x^{2} + y^{2} = 16

(c) C = (-1,-2 )& r = \(\frac{\text { diameter }}{2}=\frac{25}{2}\)

The equation of the circle is

(x + 1)^{2} + (y + 2)^{2} = \(\left(\frac{25}{2}\right)^{2}\)

x^{2} + 2x + 1 + y^{2} + 4y +4 = \(\frac { 625 }{ 4 }\)

⇒ 4x^{2} + 4y^{2} + 8x + 16y – 605 = 0.

(d) C = (1, 1) r = \(\sqrt{2}\) units

Equation of the circle is

(x – 1)^{2} +(y – 1)^{2} = \((\sqrt{2})^{2}\)

⇒ x^{2} + 1 – 2x + y^{2} + 1 – 2y = 2

⇒ x^{2} + y^{2} – 2x – 2y = 0

(e) C = (a cos θ, a sin θ) and r = a units

Equation of the circle is (x – a cos θ)^{2}+ (y – a sin θ)^{2} = a^{2}

⇒ x^{2} + a^{2} cos^{2} θ – 2a x cos θ + y^{2} + a^{2} sin^{2} θ – 2a y sin θ = a^{2}

⇒ x^{2} + y^{2} – 2a x cos θ – 2 ay sin θ + a^{2} (cos^{2} θ + sin^{2} θ) = a^{2}

⇒ x^{2} + y^{2} – 2a x cos θ – 2a y sin θ + a^{2} – a^{2}= θ

⇒ x^{2} + y^{2} – 2a x cos θ – 2a y sin θ = 0 (∵ cos^{2} θ+ sin^{2} θ = 1)

Question 2.

Find the centre of the circle, two of the diameters are

(a) x + y = 2, x – y = 0

(b) 2x – 3y = 1, 3x – 2y = 2

(c) y = 0 and y = x – 5

Answer:

(a) Given x + y = 2, x – y = 0 solving these two equations we get

(b) Given 2x – 3y = 1 → × 3

3x – 2y = 2 → × 2

(c) y = 0 ⇒ 0 = x – 5 ⇒ x = 5

∴ centre = (5,0).

Question 3.

Write the equation of the point circle with centre at

(a) (4, -5)

(b) (-3, 2)

(c) (1,0).

Answer:

(a) Point circle means circle whose radius = 0

Given C = (4,-5) & r = 0

The point circle & (x – 4)^{2} + (y + 5)^{2} = 0

⇒ x^{2} + y^{2} – 8x + 10y + 41 = 0

(b) Centre (-3,2)

The point circle is (x + 3)^{2} + (y – 2)^{2} = 0

⇒ x^{2} + y^{2} + 6x – 4y + 13 = 0

(C) Centre (1,0)

∴ Circle is (x – 1)^{2} + (y – 0)^{2} = 0

x^{2} – 2x + 1 + y^{2} = 0 ⇒ x^{2} + y^{2} – 2x + 1 = 0.

Part -B

**2nd PUC Basic Maths Circles Ex 15.1 Two Marks Questions and Answers**

Question 1.

Find the equation of the circle.

(a) Two of the diameters are x + y = 6 and x + 2y = 4 and its radius is 10 units.

(b) Two of the diameters x + y = 4 and x – y = 2 and passing through the point (2,-1).

(c) Two of the diameters are 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 Sq. cm.

Answer:

(a) Given r = 10, & Diameters are x + y = 6 & x + 2y = 4

Solving

∴ Centre = (8,-2) & r = 10 units.

∴ Equation of the circle is (x – 8)^{2 }+ (y + 2)^{2} = 10

⇒ x^{2} + y^{2} – 16x + 4y – 32 = 0.

(b) Point of inter section of two diameters x + y = 4 & x – y = 2 is the centre.

∴ Centre C = (3, 1). Also given P(2, – 1)

r = CP = \(\sqrt{(2-3)^{2}-(-1-1)^{2}}=\sqrt{(-1)^{2}+(-2)^{2}}=\sqrt{1+4}=\sqrt{5}\)

∴ The required equation of the circle with centre (3, 1) & radius = \(\sqrt{5}\) is

(x – 3)^{2} – (y – 1)^{2} = \((\sqrt{5})^{2}\)

x^{2} + 9 – 6x + y^{2} + 1 – 2y = 5

⇒ x^{2} + y^{2} – 6x – 2y + 5 = 0.

(c) Point of intersection of the two diameters is centre

2x – 3y = 5 → × 3

3x – 4y = 7 → × 2

∴ C = (1, – 1)

Also Given Area of the circle = 154 sq. cm

⇒ πr^{2} = 154

\(r^{2}=\frac{154}{\pi}=\frac{154}{22} \times 7=49\)

⇒ r = 7

∴ Required Equation of the circle with centre (1, -1) & r = 7 units is

(x – 1)^{2} + (y + 1)^{2} = 49

x^{2} + 1 – 2x + y^{2} + 2y + 1 = 49

x^{2} + y^{2} – 2x + 2y – 47 = 0.

Question 2.

Find the equation of the circle with centre at (-2, 1) and passing through the origin.

Answer:

Given C = (-2, 1) & Let P = (0,0)

r = CP = \(\sqrt{(0-(-2))^{2}+(0-1)^{2}}=\sqrt{2^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}\)

∴ Required equation of the circle with centre C = (-2, 1) & r = \(\sqrt{5}\) units is

(x + 2)^{2} + (y – 1)^{2} = \((\sqrt{5})^{2}\)

⇒ x^{2} + y^{2} + 4x – 2y = 0.

Question 3.

Find the equation of the circle with centre at (2, 1) and passing through (0, -1).

Answer:

Given Centre = C (2, 1) & Let P = (0, -1)

Here r = CP = \(\sqrt{(0-2)^{2}+(-1-1)^{2}}=\sqrt{4+4}=\sqrt{8}\) units

Equation of the circle with centre (2, 1) & r = \(\sqrt{8}\) is

(x – 2)^{2} + (y – 1)^{2} = \((\sqrt{8})^{2}\)

x^{2} + 4x + y^{2} + 1 – 2y = 8

x^{2} + y^{2} – 4x – 2y – 3 = 0.

Question 4.

Find the equation of the circle described on the line joining the points A and B as diameter where.

(a) A(-5, 1) and B(1,3)

(b) A(2,0) and B(0, 2)

(c) A(3,4) and B(1, -2)

Answer:

(a)

Equation of the circle is (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

i.e., (x + 5) (x – 1) + (y – 1) (y – 3) = 0

⇒ x^{2} + 5x – x – 5 + y^{2} – 3y – y + 3 = 0

⇒ x^{2} + y^{2} + 4x – 4y – 2 = 0.

(b)

Equation of the circle is (x – x_{1})(x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

i.e., (x – 2)(x – 0) + (y – 0) (y – 2) = 0

x^{2} – 2x + y^{2} – 2y = 0

⇒ x^{2} + y^{2} – 2x – 2y = 0

(c) Given A(3,4) B(1, -2)

∴ Required equation is (x – x_{1})(x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

i.e., (x – 3) (x – 1) + (y – 4) (y + 2) = 0

x^{2} – x – 3x + 3 + y^{2} + 2y – 4y – 8 = 0

⇒ x^{2} + y^{2} – 4x – 2y – 5 = 0.