Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(a) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(a)

I.

Question 1.

Find the equation of the circle with centre C and radius r where.

i) C = (2, -3), r = 4

Solution:

Equation of the circle is

⇒ (x – h)² + (y – k)² = a²

⇒ (x – 2)² + (y + 3)² = 4²

x² – 4x + 4 + y² + 6y + 9 = 16

x² + y² – 4x + 6y – 3 = 0

ii) C = (-1, 2), r = 5

Solution:

Equation of the circle is

(x + 1)² + (y – 2)² = 5²

⇒ x² + 2x + 1 + y² – 4y + 4 = 25

⇒ x² + y² + 2x – 4y – 20 = 0

iii) C = (a, -b); r = a + b

Solution:

Equation of the circle is

(x – a)² + (y + b)² = r²

⇒ x² – 2xa + a² + y² + 2by + b² = (a + b)²

⇒ x² + y² – 2xa + 2by – 2ab = 0

iv) C = (-a, -b); r = \(\sqrt{a^{2}-b^{2}}=2\) (|a| > |b|)

Solution:

Equation of the circle is

(x + a)² + (y + b)² = [latex]\sqrt{a^{2}-b^{2}}=2[/latex]²

⇒ x² + y² + 2xa + 2yb + a² + b² = a² – b²

⇒ x² + y² + 2xa + 2yb + 2b² = 0

v) C = (cos α, sin α); r = 1

Solution:

Equation of the circle is

(x – cos α)² + (y – sin α)² = 1

x² + y² – 2x cos α – 2y sin α + sin² α + cos² α = 1

x² + y² – 2x cos α – 2y sin α = 0

vi) C = (-7, -3);r = 4

Solution:

Equation of the circle is

(x + 7)² + (y + 3)² = 4²

x² + y² + 14x + 6y +49 + 9=16

⇒ x² + y² + 14x + 6y + 42 = 0

vii) C = (-\(\frac{1}{2}\), -9), r = 5

Solution:

Equation of the circle is

(x + \(\frac{1}{2}\))² + (y + 9)² = 5²

x² + x + \(\frac{1}{4}\) + y² + 18y + 81 =25

x² + y² + x + 18y + 56 + \(\frac{1}{4}\) = 0

4x² + 4y² + 4x + 72y + 225 = 0

viii) C = (\(\frac{5}{2}\), \(\frac{4}{3}\)), r = 6

Solution:

Equation of the circle is

(x – \(\frac{5}{2}\))² + (y + \(\frac{4}{3}\))² = 6²

Multiplying with 36

36x² + 36y² – 180x + 96y + 225 + 64 – 1296 = 0

⇒ 36x² + 36y² – 180x + 96y – 1007 = 0

ix) C = (1, 7), r = \(\frac{5}{2}\)

Solution:

Equation of the circle is

(x – 1)² + (y – 7)² = (\(\frac{5}{2}\))²

⇒ x² – 2x + 1 + y² – 14y + 49 = \(\frac{25}{4}\)

⇒ x² + y² – 2x – 14y + \(\frac{175}{4}\) = 0

4x² + 4y² – 8x – 56y + 175 = 0

x) C = (0, 0); r = 9

Solution:

Equation of the circle is

(x – 0)² + (y – 0)² = (9)²

x² + y² = 81

Question 2.

Find the equation of the circle passing through the origin and having the centre at (-4, -3).

Solution:

Equation of the circle is

(x – h)² + (y – k)² = r²; (h, k) = (-4, -3)

(x + 4)² + (y + 3)² = r²

Circle passes through origin.

∴ (0 + 4)² + (0 + 3)² = r² ⇒ 25 = r²

Then required equation of circle be

(x + 4)² + (y + 3)² = 25

x² + y² + 8x + 6y = 0

Question 3.

Find the equation of the circle passing through (2, -1) having the centre at (2, 3).

Solution:

C = (2, 3), P = (2, -1)

Radius CP = \(\sqrt{(2 – 2)^{2} + (3 + 1)^{2}}\) = 4

Equation of circle be

(x – 2)² + (y – 3)² = 42

x² + y² – 4x – 6y – 3 = 0

Question 4.

Find the equation of the circle passing the through (- 2, 3) centre at (0, 0).

Solution:

C = (0, 0), P = (- 2, 3)

Radius = \(\sqrt{(0 + 2)^{2} + (0 – 3)^{2}}\)

= √13

Equation of circle be

(x – 0)² + (y – 0)² = (√13)²

x² + y² = 13

Question 5.

Find the equation of the circle passing through (3, 4) having and the centre at (-3, 4).

Solution:

Let the equation of the circle be

(x – h)² + (y – k)² – r²

Centre (h, k) = (-3, 4)

(x + 3)² + (y – 4)² = r²

Circle passes through (3, 4)

(3 + 3)² + (4 – 4)² = r²

r² = 36

Equation of the circle is

(x + 3)² + (y – 4)² = 36

x² + 6x + 9 + y² – 8y + 18 – 36 = 0

x² + y² + 6x – 8y – 11 =0

Question 6.

Find the value of ‘a’ if 2x² + ay² – 3x + 2y – 1 =0 represents a circle and also find its radius.

Solution:

General equation of second degree

ax² + 2hxy + by² + 2gx + 2fy + c = 0

Represents a circle, when

a = b, h= 0, g² + f² – c > 0 In

2x² + ay² – 3x + 2y -1 = 0

a = 2, above equation represents circle.

x² + y² – \(\frac{3}{2}\) x + y – \(\frac{1}{2}\) = 0

2g = –\(\frac{3}{2}\); 2f = 1; c = –\(\frac{1}{2}\)

c = (-g, -f) = (\(\frac{+3}{4}\), \(\frac{-1}{2}\))

Radius = \(\sqrt{g^{2} + f^{2} – c}\) = \(\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}\)

= \(\frac{\sqrt{21}}{4}\) units

Question 7.

Find the values of a, b if ax² + bxy + 3y² – 5x + 2y – 3 = 0 represents a circle. Also find the radius and centre of the circle.

Solution:

General equation of second degree

ax² + 2hxy + by² + 2gx + 2fy + c = 0

represents a circle a = b, h = 0

∴ ax² + bxy + 3y² – 5x + 2y – 3 = 0

represents, a circle.

When b = 0, a = 3

3x² + 3y² – 5x + 2y – 3 = 0

Question 8.

If x² + y² + 2gx + 2fy -12 = 0 represents a circle with centre (2, 3), find g, f and its radius.

Solution:

Circle is x² + y² + 2gx + 2fy – 12 = 0

C = (-g, -f) C = (2, 3)

∴ g = -2, f = -3, c = – 12

Radius = \(\sqrt{g^{2} + f^{2} – c}\)

= \(\sqrt{4 + 9 + 12}\)

= 5 units

Question 9.

If x² + y² + 2gx + 2fy = 0 represents a circle with Centre (-4, -3) then find g,f and the radius of the circle.

Solution:

Circle is x² + y² + 2gx + 2fy – 12 = 0

⇒ C = (-g, -f) C = (-4, -3)

∴ g = 4, f = 3

Radius = \(\sqrt{g^{2} + f^{2} – c}\)

= \(\sqrt{16 + 9}\) = 5 units

Question 10.

If x² + y² – 4x + 6y + c = 0 represents a circle with radius 6 then find the value of c.

Solution:

Circle is x² + y² – 4x + 6y + c = 0

r = \(\sqrt{g^{2} + f^{2} – c}\) ; g = – 2, f = 3

6 = \(\sqrt{4 + 9 – c}\)

36 = 13 – c

or c = -23

Question 11.

Find the centre and radius of each of the circles whose equations are given below.

i) x² + y² – 4x – 8y – 41 = 0

Solution:

x² + y² – 4x – 8y – 41 = 0 ……….. (i)

x² + y² + 2gx + 2fy + c = 0 ……….. (ii)

Comparing (i) and (ii) we get

g = – 2, f = – 4, c = – 41

Radius = \(\sqrt{g^{2} + f^{2} – c}\)

= \(\sqrt{4 + 16 + 41}\)

= \(\sqrt{61}\) units

Centre = (-g, -f) = (2, 4)

ii) 3x² + 3y² – 5x- 6y + 4 = 0

Solution:

Equation of the circle is

3x² + 3y² – 5x – 6y + 4 = 0

x² + y² – \(\frac{5}{3}\)x – \(\frac{6}{3}\)y + \(\frac{4}{3}\) = 0 …………. (i)

x² + y² + 2gx + 2fy + c = 0 …………. (ii)

Comparing equations (i) and (ii) we get

iii) 3x² + 3y² + 6x – 12y – 1 = 0

Find the radius and centre of the circle.

Solution:

Equation of the circle is

3x² + 3y² + 6x – 12y – 1 = 0

x² + y² + \(\frac{6}{3}\)x – \(\frac{12}{3}\)y – \(\frac{1}{3}\) = 0

C = (-g, -f) = (-1, 2)

Radius = \(\sqrt{g^{2} + f^{2} – c}\)

= \(\sqrt{1 + 4 + \frac{1}{3}}\)

= \(\frac{4}{\sqrt{3}}\)

iv) x² + y² + 6x + 8y – 96 = 0

Solution:

Equation of the circle is

x² + y² + 6x + 8y – 96 = 0

C = (-g, -f) = (-3, -4)

Radius = \(\sqrt{g^{2} + f^{2} – c}\)

= \(\sqrt{9 + 16 + 96}\)

= \(\sqrt{121}\) = 11 units

v) 2x² + 2y² – 4x + 6y – 3 = 0 Sol. Equation of the circle is

Solution:

x² + y² – 2x + 3y-|- = 0 ………….. (i)

x² + y² + 2gx + 2fy + c = 0 …………… (ii)

Comparing (i) and (ii) we get

C = (1, –\(\frac{3}{2}\))

Radius = \(\sqrt{g^{2} + f^{2} – c}\)

= \(\sqrt{1 + \frac{9}{4} + \frac{3}{2}}=\frac{\sqrt{19}}{2}\) units

vi) 2x² + 2y² – 3x + 2y – 1 = 0

Solution:

Equation of the circle is

x + y – \(\frac{3}{2}\)x + y – \(\frac{1}{2}\) = 0

x² + y² + 2gx + 2fy + c = 0

Comparing we get

C = (-g, -f) = (\(\frac{3}{2}\), –\(\frac{1}{2}\))

r = \(\sqrt{g^{2} + f^{2} – c}\)

\(\sqrt{\frac{9}{16} + \frac{1}{4} + \frac{3}{2}}=\frac{\sqrt{21}}{4}\) units

vii) \(\sqrt{1 + m^{2}}\) (x² + y²) – 2cx – 2mcy = 0

Solution:

Equation of the circle is

viii) x² + y² + 2ax – 2by + b2 = 0

Solution:

Equation of the circle is

x² + y² + 2gx + 2fy + c = 0

C = (-g, -f) = (-a, b)

Radius = \(\sqrt{g^{2} + f^{2} – c}\)

= \(\sqrt{a^{2} + b^{2} – b^{2}}\) = a units

Question 12.

Find the equations of the circles for which the points given below are the end points of a diameter.

i) (1, 2), (4, 6)

Solution:

Equation of the circle with (x_{1}, y_{1}), (x_{2}, y_{2}) as ends of a diameter is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x – 1) (x – 4) + (y – 2) (y – 6) = 0

⇒ x² – 5x + 4 + y² – 8y + 12 = 0

⇒ x² + y² – 5x – 8y + 16 = 0

ii) (-4, 3); (3, -4)

Solution:

Equation of circle with (x_{1}, y_{1}) and (x_{2}, y_{2}) are end points of diameter is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = o

Required equation of circle be

(x + 4) (x – 3) + (y – 3) ( y + 4) = 0

x² + y² + x + y – 24 = 0

iii) (1, 2); (8, 6)

Solution:

Equation of the circle is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 1) (x – 8) + (y – 2) (y – 6) = 0

x² + y² – 9x – 8y + 20 = 0

iv) (4, 2); (1, 5)

Solution:

Equation of the circle is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x – 4) (x – 1) + (y – 2) (y – 5) = 0

x² + y² – 5x – 7y + 14 = 0

v) (7, -3); (3, 5)

Solution:

Equation of the circle is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x – 7)(x – 3) + (y + 3) (y – 5) = 0

x² + y² – 10x – 2y + 6 = 0

vi) (1, 1); (2, -1)

Solution:

Equation of the circle is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x – 1) (x – 2) + (y – 1) (y + 1) = 0

x² + y² – 3x +1=0

vii) (0, 0); (8, 5)

Solution:

Equation of the circle is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x – 0) (x – 8) + (y – 0) (y – 5) = 0

⇒ x² – 8x + y² – 5y = 0

x² + y² – 8x – 5y = 0

viii) (3, 1); (2, 7)

Solution:

Equation of the circle is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 3) (x – 2) + (y – 1) (y – 7) = 0

x²+ y² – 5x – 8y + 13 = 0

Question 13.

Obtain the parametric equation of each of the following circles.

i) x² + y² = 4

Solution:

Equation of the circle is x² + y² = 4

C (0, 0), r = 2

Parametric equations are

x = – g + r cos θ = 2 cos θ

y = – b + r sin θ = 2 sin θ, 0 ≤ θ < 2π

ii) 4(x² + y²) = 9

Solution:

Equation of the circle is 4(x² + y²) = 9

x² + y² = \(\frac{9}{4}\)

C(0, 0), r = \(\frac{3}{2}\)

Parametric equations are

x = \(\frac{3}{2}\) cos θ, y = \(\frac{3}{2}\) sin θ, 0 ≤ θ < 2π

iii) 2x² + 2y² = 7

Solution:

Equation of the circle is 2x² + 2y² = 7

x² + y² = \(\frac{7}{2}\)

C(0, 0), r = \(\sqrt{\frac{7}{2}}\)

Parametric equations are

x = \(\sqrt{\frac{7}{2}}\) cos θ, y = \(\sqrt{\frac{7}{2}}\) sin θ, 0 ≤ θ < 2π

iv) (x – 3)² + (y – 4)² = 8²

Solution:

Equation of the circle is (x – 3)² + (y – 4)² = 8²

Centre (3, 4), r = 8

Parametric equations are

x = 3 + 8 cos θ, y = 4 + 8 sin θ, 0 ≤ θ < 2π

v) x² + y² – 4x – 6y – 12 = 0

Solution:

Equation of the circle is

x² + y² – 4x – 6y – 12 = 0

Centre (2, 3), r = \(\sqrt{4 + 9 + 12}\) = 5

Parametric equations are

x = 2 + 5 cos θ, y = 3 + 5 sin θ, 0 ≤ θ < 2π

vi) x² + y² – 6x + 4y – 12 = 0

Solution:

Equation of the circle is x² + y² – 6x + 4y – 12 = 0

Centre (3, – 2), r = \(\sqrt{9 + 4 + 12}\) = 5

Parametric equations are

x = 3 + 5 cos θ, y = -2 + 5 sin θ, 0 ≤ θ < 2π

II.

Question 1.

If the abscissae of points A, B are the roots of the equation, x² + 2ax – b² = 0 and ordinates of A, B are roots of y² + 2py – q²= 0, then find the equation of a circle for which \(\overline{\mathrm{AB}}\) is a diameter.

Solution:

Equation of the circle is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

x² – x(x_{1} + x_{2}) + x_{1}x_{2} + y² – y (y_{1} + y_{2}) + y_{1}y_{2} = 0

x_{1}, x_{2} are roots of x² + 2ax – b² = 0

y_{1}, y_{2} are roots of y² + 2py – q² = 0

x_{1}x_{2} = -2a, y_{1}y_{2} = -2p

x_{1}x_{2} = -b², y_{1}y_{2} = -q²

Equation of circle be

x² – x(- 2a) – b² + y² – y(- 2p) – q² = 0

x² + 2xa + y² + 2py – b² – q² = 0

Question 2.

i) Show that A (3, -1) lies on the circle x² + y² – 2x + 4y = 0. Also find the other end of the diameter through A.

Solution:

Equation of the circle is

x² + y² – 2x + 4y = 0 …………… (i)

A (3, -1); Let B (x_{1}, y_{1})

Substituting A in equation (i)

(3)² + (-1)² – 2(3) + 4 (-1) = 0

∴ A lies on the circle

C (-g, -f)

C = (1, -2)

C is the centre of circle

C is midpoint of AB

\(\frac{x_{1}+3}{2}\) = -1 \(\frac{y_{1}-1}{2}\) = -2

x_{1} = -1 y_{1} = -3

B(x_{1}, y_{1}) = (-1, -3)

ii) Show that A(-3, 0) lies on x² + y² + 8x + 12y + 15 = 0 and find the other end of diameter through A.

Solution:

If A (-3, 0) satisfy

x² + y² + 8x + 12y + 15 = 0

then (-3)² + (0)² – 8 × 3 + 12 × 0 +15

9 – 24 + 1 5 = 0

∴ (-3, 0) is one end of diameter.

A (-3, 0), C(-4, -6), B(x_{1}, y_{1})

\(\frac{x_{1}+(-3)}{2}\) = -4 \(\frac{y_{1}+0}{2}\) = -6

x_{1} = -5 y_{1} = -12

∴ Other end of diameter is (-5, -12)

Question 3.

Find the equation of a circle which passes through (2, -3) and (-4, 5) and having the centre on 4x + 3y + 1 =0

Solution:

x² + y² +2gx + 2fy + c = 0 ……….. (i)

Equation (i) passes through (2, -3), (-4, 5)

∴ 4 + 9 + 4g – 6f + c = 0 …………. (ii)

16 + 25 – 8g + 10f + c = 0 …………. (iii)

Equation (iii) – (ii) we get

28 – 12g + 16f = 0

(or) 3g – 4f = 7

Centre lies on (-g, -f) lies on 4x + 3y + 1 = 0

then 4(-g) + 3(-f) + 1 = 0

3g – 4f – 7 = 0

Solving we get f = -1

g = 1

Now substituting f, g values in equation (ii) we get

4 + 9 + 4(1) – 6 (-1) + c = 0, c = -23

x² + y² + 2x – 2y – 23 = 0 is required equation of circle.

Question 4.

Find the equation of a circle which passes through (4, 1) (6, 5) and having the centre on 4x + 3y – 24 = 0

Solution:

Equation of circle be

x² + y² + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then

4² + 1² + 2g(4) + 2f(1) + c = 0 ………… (i)

6² + 5² + 2g(6) + 2f(5) + c = 0 ………… (ii)

Centre lie on 4x + 3y – 24 = 0

∴ 4(-g) + 3 (-f) – 24 = 0 ………… (iii)

(ii) – (i) we get

44 + 4g + 8f = 0 …………… (iv)

Solving (iii) and (iv) we get

f = -4, g = -3, c = 15

∴ Required equation of circle is

x² + y² – 6x – 8y +15 = 0

Question 5.

Find the equation of a circle which is concentric with x² + y² – 6x – 4y – 12 = 0 and passing through (-2, 14).

Solution:

x² + y² – 6x – 4y – 12 = 0 ……… (i)

C = (-g, -f)

= (3, 2)

Equation of circle concentric with (i) be

(x – 3)² + (y – 2)² = r²

Passes through (-2, 14)

∴ (-2 – 3)² + (14 – 2)² = r²

169 = r²

Required equation of circle be

(x – 3)² + (y – 2)² = 169

x² + y² – 6x – 4y – 156 = 0

Question 6.

Find the equation of the circle whose centre lies on the X – axis and passing through (-2, 3) and (4, 5).

Solution:

x² + y² + 2gx + 2fy + c = 0 ………… (i)

(-2, 3) and (4, 5) passes through (i)

4 + 9 – 4g + 6f + c = 0 ………….. (ii)

16 + 25 + ,8g + 10f + c = 0 ………… (iii)

(iii) – (ii) we get

28 + 12g + 4f = 0

f + 3g = -7

Centre lies on X – axis then f = 0

g = -, \(\frac{7}{3}\) f = 0, c = \(\frac{67}{3}\) -, we get by substituting g, f in equation (ii)

Required equation will be

3(x² + y²) – 14x – 67 = 0

Question 7.

If ABCD is a square then show that the points A, B, C and D are concyclic.

Solution:

AB = a, AD = a

A (0, 0), B(0, a), D (a, 0)

Equation of circle is

x² + y² + 2gx + 2fy + c = 0

passes through A, B, D we get

A : 0 + 0 + 2g(0) + 2f(0) + c = 0

C = 0

B : 0 + a2 + 2g(0) + 2fa + 0 = 0

f = \(\frac{a}{2}\)

Similarly g = –\(\frac{a}{2}\)

Required equation of circle be

x² + y² – ax – ay = 0

Co-ordinates of C are (a, a)

a² + a² – a² – a² = 0

⇒ C lies on the circle passing through A, B, D.

∴ A, B, C, D are concyclic.

III.

Question 1.

Find the equation of circle passing through each of the following three points.

i) (3, 4); (3, 2); (1, 4)

Solution:

Equation of circle

x² + y² + 2gx + 2fy + c = 0

Given points satisfy above equation then

9 +16 + 6g + 8f + c + = 0 ………….. (i)

9 + 4 + 6g + 4f + c = 0 ………….. (ii)

1 + 16 + 2g + 8f + c = 0 ………… (iii)

Subtracting (ii) – (i) we get

-12 – 4f = 0 (or) f = – 3

(ii) – (iii) we get

– 4 + 4g – 4f = 0

g – f = 1

9 = – 2

Now substituting g, f in equation (i) we get

25 + 6 (- 2) + 8 (- 3) + c = 0

we get c = 11

Required equation of circle be

x² + y² – 4x – 6y + 11 = 0

ii) (1, 2); (3, -4); (5, -6)

Solution:

Equation of circle is

x² + y² + 2gx + 2fy + c = 0

1 + 4 + 2g + 4f + c = 0 …………. (i)

9 + 16 + 6g — 8f + c = 0 …………. (ii)

25 + 36 + 10g – 12f + c = 0 …………. (iii)

Subtracting (ii) – (i) we get

20 + 4g – 12f = 0

(or) 5 + g – 3f = 0 …………. (iv)

Similarly (iii) – (ii), we get

36 + 4g – 4f = 0

(or) 9 + g – f = 0 …………. (v)

Solving (v) and (iv) we get

f = -2, g = -11, c = 25

Required equation of circle be

x² + y² – 22x – 4y + 25 = 0

iii) (2, 1); (5, 5); (-6, 7)

Solution:

Equation of circle is

x² + y² + 2gx + 2fy + c = 0

4 + 1 + 4g + 2f + c = 0 …………. (i)

25 + 25 + 10g + 10f + c = 0 …………. (ii)

36 + 49 – 12g + 14f + c = 0 …………. (iii)

Subtracting (ii) – (i)

45 + 6g + 8f = 0 …………. (iv)

Subtracting (iii) – (ii)

35 – 22g + 4f = 0 …………. (v)

Solving (iv) and (v) we get

g = \(\frac{1}{2}\); f = -6; c = 5

Required equation of circle be

x² + y² + x – 12y + 5 = 0

iv) (5, 7); (8, 1); (1, 3)

Solution:

Equation of circle is

x² + y² + 2gx + 2fy + c = 0

25 + 49 + 10g + 14f + c = 0 …………. (i)

64 + 1 + 16g + 2f + c = 0 …………. (ii)

1 + 9 + 2g + 6f + c = 0 …………. (iii)

Subtracting (ii) – (i)

-9 + 6g – 12f = 0 …………. (iv)

(or) 2g – 4f – 3 = 0

Subtracting (iii) – (ii)

– 55 – 14g + 4f = 0 ……….. (v)

Solving (v) and (iv) we get

g = \(\frac{-29}{6}\), f = \(\frac{-19}{6}\), c = \(\frac{56}{3}\)

∴ Required equation of circle is

x² + y² – \(\frac{-29}{3}\)x – \(\frac{-19}{3}\)y + \(\frac{56}{3}\) = 0

3(x² + y²) – 29x – 19y + 56 = 0

Question 2.

i) Find the equation of the circle passing through (0, 0) and making intercepts 4,3 on X- axis and Y – axis respectively.

Solution:

x² + y² + 2gx + 2fy + c = 0

Circle is passing through

(0, 0), (4, 0) and (0, 3)

0 + 0 + 2g(0) + 2f(0) + c = 0

c = 0 ………… (i)

16 + 0 + 8g + 2f . 0 + c = 0

g = -2 as c = 0

Similarly 0 + 9 + 2g. 0 + 6f + c = 0

f = –\(\frac{3}{2}\) as c = 0

Required equation of circle will be

x² + y² – 4x – 3y = 0

If intercepts are negative when circle passes through the points (0, 0) (-4, 0) (0, -3). Required equation of circle is x² + y² + 4x + 3y = 0

ii) Find the equation of the circle passing through (0, 0) and making intercept 6 units on X – axis and intercept 4 units on Y – axis.

Solution:

OA = 6 units,

OB = 4 units,

OD = 3 units, OE = 2 units.

∴ Co-ordinates of centre b(3, 2)

Radius OC = \(\sqrt{(0+3)^{2}+(0-2)^{2}}\) = √13

Equation of circle with (h, k) as centre r be radius is (x – h)² + (y – k)² = r²

∴ Required equation of circle be

(x – 3)² + (y – 2)² = 13

x² + y² – 6x – 4y = 0

If intercepts are negative when circle passes through the points (0, 0) (-6, 0) (0, -4)

Required equation of circle is x² + y² + 6x + 4y = 0

Question 3.

Show that the following four points in each of the following are concyclic and find the equation of the circle on which they lie.

i) (1, 1), (-6, 0), (-2, 2), (-2, -8)

Solution:

Suppose the equation of the required circle is

x² + y² + 2gx + 2fy + c = 0 ………… (i)

This circle passes through A (1, 1)

1 + 1 + 2g + 2f + c = 0

⇒ 2g + 2f + c = -2 …………. (ii)

The circle passes through B (-6, 0)

36 + 0 – 12g + 0 + c = 0

– 12g + c = – 36 …………. (iii)

This circle passes through C (-2, 2)

4 + 4 – 4g + 4f + c = 0

– 4g + 4f + c = – 8 ………… (iv)

(iii) – (iv) gives – 8g – 4f = 0

⇒ 2g + f = 7

(i) – (ii) gives 14g + 2f = 34

7g + f = 17 ………….. (v)

Solving (iv) and (v) we get g = 2, f = 3

Putting g = 2, f = 3 in

4 + 6 + c = -2

c = – 12

Required circle is x² + y² + 4x + 6y – 12 = 0

Substituting (-2, -8) in this above equation, we get

4 + 64 – 8- 48 -12 = 68 – 68 = 0

(- 2, -8) satisfies the above equation

∴ A, B, C, D are concyclic.

Equation of the circle is

x² + y² + 4x + 6y – 12 = 0

ii) (1, 2); (3, -4); (5, -6); (19, 8)

Solution:

Equation of circle

x² + y² + 2gx + 2fy + c = 0

1 + 4 + 2g + 4f + c = 0 …………. (i)

9 + 16 + 6g – 8f + c = 0 …………. (ii)

25 + 36 + 10g – 12f + c = 0 …………. (iii)

Subtracting (ii) – (i) we get

20 + 4g – 12f = 0

5 + g – 3f = 0 …………. (iv)

Subtracting (iii) – (ii) we get

36 + 4g – 4f = 0

(or)

9 + g – f = 0 …………. (v)

Solving (iv) and (v) we get

f = -2, g = – 11, c = 25

Equation of circle will be

x² + y² – 22x – 4y + 25 = 0 ……… (vi)

(19, 8) substituting in equation (vi)

(19)² + 8² – 22 × 19 – 4 × 8 + 25 = 0

Hence (19, 8) lie on circle and four points are concyclic.

iii) (1, -6); (5, 2); (7, 0); (-1, -4)

Solution:

Equation of circle

x² + y² + 2gx + 2fy + c =0

1 + 36 + 2g- 12f + c = 0 ……….. (i)

25 + 4 + 10g + 4f + c = 0 …………. (ii)

49 + 14g + c = 0 …………… (iii)

Subtracting (ii) – (i) we get

-8 + 8g + 16f = 0

(or)

2f+ g – 1 = 0 ……… (iv)

Subtracting (iii) – (ii) we get

20 + 4g – 4f = 0 ………… (v)

(or)

5 + g – f = 0

Solving (iv) and (v) we get

f = 2, g = -3, c = -7

Equation of circle be

x² + y² – 6x + 4y- 7 = 0 …….. (vi)

(-1, -4) satisfies equation (vi)

∴ Four points are concyclic.

iv) (9, 1), (7, 9), (-2, 12), (6, 10)

Solution:

Equation of the circle is

x² + y² + 2gx + 2fy + c = 0

This circle passes through

A(9, 1), B(7, 9), C (-2, 12)

81 + 1 + 18g + 2f + c = 0 ………….. (i)

49 + 81 14g + 18f + c = 0 ………….. (ii)

4 + 144 – 4g + 24f + c = 0 ………….. (iii)

(ii) – (i) gives – 4g + 16f + 48 = 0

4g – 16f = 48

g – 4f = 12 ……….. (iv)

(ii) – (iii) gives 18g – 6f – 18 = 0

18g – 6f = 18 ………… (v)

36g – 12f = 36 ……….. (v) × 2

3g – 12f = 36 ………… (iv) × 3

Subtracting 33g = 0 ⇒ g = 0

Substituting in (iv) we get – 4f = 12

f = \(\frac{12}{-4}\) = -3

Substituting the values of g, f in (i)

18(0) + 2 (-3) + c + 82 = 0

c = 6 – 82 = -76

Equation of the required circle is

x² + y² – 6y – 76 = 0

x² + y² – 6y – 76 = 6² + 10² – 6(10) – 76

= 36 + 100 – 60 – 76

= 136 – 136 = 0

D(6, 10) lies on the circle passing through A, B, C.

∴ A, B, C and D are concylic.

Equation of the circle is x² + y² – 6y – 76 = 0

Question 4.

If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic, and then find c.

Solution:

x² + y² + 2gx + 2fy + c = 0

Satisfies (2, 0), (0, 1) (4, 5) we get

4 + 0 + 4g + c_{1} = 0 ………… (i)

0 + 1 + 2g. 0 + 2f + c_{1} = 0 ………….. (ii)

16 + 25 + 8g + 10f + c_{1} = 0 ……….. (iii)

(ii) – (i) we get

-3 – 4g + 2f = 0

4g – 2f = -3 ………. (iv)

(ii) – (iii) we get

-40 – 8g – 8f =0 (or)

g + f = – 5 ……… (v)

Solving(iv) and (v) we get

g = –\(\frac{13}{6}\), f = \(\frac{17}{6}\)

Substituting g and f values in equation (i) we get

4 + 4(-\(\frac{13}{6}\)) + c_{1} = 0

c_{1} = \(\frac{14}{3}\)

Now equation x² + y² – \(\frac{13}{3}\)x – \(\frac{17}{3}\)y + \(\frac{14}{3}\) = 0

Now circle passes through (0, c) then

c² – \(\frac{17}{3}\)c + \(\frac{14}{3}\)

3c² – 17c + 14 = 0

⇒ (3c – 14) (c – 1) = 0

(or)

c = 1 or \(\frac{14}{3}\).

Question 5.

Find the equation of the circum circle of the triangle formed by the straight lines given in each of the following:

i) 2x + y = 4; x + y = 6; x + 2y = 5

Solution:

AB: 2x + y = 4, AB : 2x + y = 4

BC : x + y = 6, AC : x + 2y = 5

B : (-2, 8), A : (1, 2)

AC: x + 2y = 5

BC : x + y = 6

C : (7, -1)

Equation of circle

x² + y² + 2gx + 2fy + c = 0

passes through A, B, C

∴ 4 + 64 – 4g + 16f + c = 0 ……… (i)

1 + 4 + 2g + 4f + c = 0 ………… (ii)

49 + 1 + 14g – 2f + c = 0 ……….. (iii)

(i) – (ii) we get

21 – 2g + 4f = 0 ……….. (iv)

(iii) – (ii) we get

15 + 4g – 2f = 0 ……….. (v)

Solving (iv) and (v) we get f = – \(\frac{19}{2}\)

g = –\(\frac{17}{2}\); c = 50

We get substituting g, f in equation (i)

Required equation of circle be

x² + y² – 17x – 19y + 50 = 0

ii) x + 3y – 1 = 0; x + y + 1 = 0; 2x + 3y + 4 = 0

Solution:

AB : x + 3y – 1 = 0 AB : x + 3y – 1 = 0

BC : x + y +1 = 0 AC: 2x + 3y‘ + 4 = 0

B : (-2, 1) A: (-5, 2)

AC : 2x + 3y + 4 = 0

BC : x + y + 1 = 0

C : (1, -2)

Equation of circle x² + y² + 2gx + 2fy + c = 0

A, B, C are points on circumference.

25 + 4 – 10g + 4f + c = 0 ……….. (i)

4 + 1 – 4g + 2f + c = 0 ……… (ii)

∴ 1 + 4 + 2g – 4f + c = 0 ……… (iii)

Subtracting (iii) – (ii) we get

6g – 6f = 0 (or) g = f ……… (iv)

Subtracting (iii) – (i) we get

24 – 12g + 8f = 0 ……… (v)

Solving (iv) and (v) we get

g = 6, f = 6, c = 7

Required equation of circle be

x² + y² + 12x + 12y + 7

iii) 5x – 3y + 4 = 0; 2x + 3y – 5 = 0; x + y = 0.

Solution:

AB : 5x – 3y + 4 = 0

AC : 2x + 3y – 5 = 0

BC : x + y = 0

A : (\(\frac{1}{7}\), \(\frac{11}{7}\)) C : (-5, 5)

B : (-\(\frac{1}{2}\), \(\frac{1}{2}\))

Equation of circle is

x² + y² + 2gx + 2fy + c = 0

Points A, B, C are on circumference of circle

\(\frac{1}{49}\) + \(\frac{121}{49}\) + \(\frac{2}{7}\)g + \(\frac{22}{7}\)f + c = 0 ………. (i)

25 + 25 – 10g + 10f + c = 0 ………. (ii)

\(\frac{1}{4}\) + \(\frac{1}{4}\) – g + f + c ……….. (iii)

(Or)

1 – 2g + 2f + 2c = 0

(ii) — (iii) we get

(50 – \(\frac{1}{2}\) – 9g + 9f = 0

\(\frac{11}{2}\) – g + f = 0 ………… (iv)

(iii) – (i) we get

Solving (iv) and (v) we get

g = \(\frac{40}{14}\); f = \(\frac{-37}{14}\); h = \(\frac{70}{14}\)

x² + y² + \(\frac{80}{2}\)x + \(\frac{74}{14}\)y + \(\frac{70}{14}\) = 0

Required equation of circle be

7(x² + y²) + 40x – 37y + 35 = 0

iv) x – y – 2 = 0;

2x – 3y + 4 = 0;

3x – y + 6 = 0

Solution:

AB : x – y – 2 = 0

B : (10, 8) A: (-4, -6)

BC : 2x – 3y + 4 = 0

AC : 3x – y + 6 = 0

C : (- 2, 0)

Equation of circle is

x² + y² + 2gx + 2fy + c = 0

Points A, B, C are on circumference of circle

100 + 64 + 20g + 16f + c = 0 ……… (i)

16 + 36 – 8g – 12f + c = 0 ……… (ii)

4 – 4g + c = 0 ……….. (iii)

Solving above equations we get

g = – 12, f = 8, c = – 52

Required equation of circle is

x² + y² – 24x + 16y – 52 = 0

Question 6.

Show that the locus of the point of intersection of the lines x cos a + y sin a = a, x sin a – y cos a = b (a is a parameter) is a circle.

Solution:

Equations of the given lines are

x cos α + y sin α = a

x sin α -y cos α = b

Let P (x_{1}, y_{1}) be the point of intersection

x_{1} cos α + y_{1} sin α = a ………. (1)

x_{1} sin α – y_{1} cos α = b ……… (2)

Squaring and adding (1) and (2)

(x_{1}cos α + y_{1}sin α)² + (x_{1}sin α – y_{1}cos α)² = a² + b²

x²_{1} cos²α + y²_{1} sin²α + 2x_{1}y_{1} cos α sin α + x²_{1} sin² α + y²_{1} cos² α – 2x_{1}y_{1} cos α sin α = a² + b²

x²_{1} (cos²α +sin²α) + y²_{1} (sin² α + cos² α) = a² + b²

x²_{1} + y²_{1} = a² + b²

Locus of P(x_{1}, y_{1}) is the circle

x² + y² = a² + b²

Question 7.

Show that the locus of a point such that the ratio of distance of it from two given points is constant k (≠ ± 1) is a circle.

Solution:

Let P(x_{1}, y_{1}) be a point on the locus

Let A (a, 0), B(-a, 0) be two given points

By squaring and cross multiplying, we get

(x_{1} – a)² + y_{1}² = k² [(x_{1} + a)2 + y_{1}2 ]

⇒ (1 – k²) (x²_{1} + y²_{1} + a²) + (-1 – k²) (2ax_{1}) = 0

⇒ x²_{1} + y²_{1} – 2 \(\frac{(1+k^{2})}{1-k^{2}}\)ax + a² = 0

∴ Locus of P(x_{1}, y_{1}) is

x² + y² – 2\(\frac{1+k^{2}}{1-k^{2}}\)ax + a² = 0

which represents a circle. (Here k ≠± 1)