Class 11: The Circle – Exercise 24.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: The general equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ where the center is $\displaystyle ( -g, -f)$ and radius $\displaystyle = \sqrt{g^2 +f^2 - c}$

Question 1: Find the coordinates of the center and radius of each of the following circles :

i) $\displaystyle x^2 + y^2 + 6x - 8 y - 24 = 0$ ii) $\displaystyle 2x^2 + 2y^2 - 3x + 5y = 7$

iii) $\displaystyle \frac{1}{2} ( x^2 + y^2 ) + x \cos \theta + y \sin \theta - 4 = 0$ iv) $\displaystyle x^2 + y^2 -ax-by = 0$

Answer:

i) Give equation : $\displaystyle x^2 + y^2 + 6x - 8 y - 24 = 0$

Comparing it with $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ , we get,

$\displaystyle g = - 3 , \hspace{0.5cm} f = - 4 , \hspace{0.5cm} c = - 24$

Therefore center $\displaystyle = ( -g, -f) = ( -3, 4)$

Radius $\displaystyle = \sqrt{g^2 +f^2 - c} = \sqrt{ 3^2 + ( -4)^2 - ( -24) } = \sqrt{49} = 7$

ii) Give equation : $\displaystyle 2x^2 + 2y^2 - 3x + 5y = 7$

$\displaystyle \Rightarrow x^2 + y^2 - \frac{3}{2} x + \frac{5}{2} y - \frac{7}{2} = 0$

Comparing it with $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ , we get,

$\displaystyle g = \frac{-3}{4} , \hspace{0.5cm} f = \frac{5}{4} , \hspace{0.5cm} c = \frac{-7}{2}$

Therefore center $\displaystyle = ( -g, -f) = ( \frac{3}{4} , - \frac{5}{4} )$

Radius $\displaystyle = \sqrt{g^2 +f^2 - c} = \sqrt{ (\frac{-3}{4})^2 + ( \frac{5}{4} )^2 - ( \frac{-7}{2}) }$

$\displaystyle = \sqrt{\frac{9}{16}+\frac{25}{16} + \frac{7}{2}} = \frac{\sqrt{90}}{4} = \frac{3\sqrt{10}}{4}$

iii) Given equation $\displaystyle \frac{1}{2} ( x^2 + y^2 ) + x \cos \theta + y \sin \theta - 4 = 0$

$\displaystyle \Rightarrow x^2 + y^2 + 2 \cos \theta x + 2 \sin \theta y - 8 = 0$

Comparing it with $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ , we get,

$\displaystyle g = \cos \theta , \hspace{0.5cm} f = \sin \theta , \hspace{0.5cm} c = - 8$

Therefore center $\displaystyle = ( -g, -f) = ( -\cos \theta, - \sin \theta)$

Radius $\displaystyle = \sqrt{g^2 +f^2 - c} = \sqrt{ (- \cos \theta)^2 + ( - \sin \theta )^2 - ( -8) } = \sqrt{9} = 3$

iv) Given equation $\displaystyle x^2 + y^2 -ax-by = 0$

Comparing it with $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ , we get,

$\displaystyle g = \frac{-a}{2} , \hspace{0.5cm} f = \frac{-b}{2} , \hspace{0.5cm} c = 0$

Therefore center $\displaystyle = ( -g, -f) = ( \frac{a}{2} , \frac{b}{2} )$

Radius $\displaystyle = \sqrt{g^2 +f^2 - c} = \sqrt{ (\frac{-a}{2} )^2 + ( \frac{-b}{2})^2 - ( 0) } = \frac{\sqrt{a^2+b^2}}{2}$

$\displaystyle \\$

Question 2: Find the equation of the circle passing through the points:

(i) $\displaystyle (5,7), (8, 1)$ and $\displaystyle (1, 3)$ (ii) $\displaystyle (1,2), (3, - 4)$ ,and $\displaystyle (5, - 6)$

(iii) $\displaystyle (5, - 8), (- 2,9)$ and $\displaystyle (2,1)$ (iv) $\displaystyle (0, 0), (-2, 1)$ and $\displaystyle (- 3,2)$

Answer:

(i) The circle passes through the points $\displaystyle (5,7), (8, 1)$ and $\displaystyle (1, 3)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle (5, 7)$ in i) we get

$\displaystyle 25 + 49 + 10g + 14f + c = 0 \hspace{0.5cm} \Rightarrow 10g+14f+c = -74$ … … … … … ii)

Substituting $\displaystyle (8, 1)$ in i) we get

$\displaystyle 64 + 1 + 16g + 2f + c = 0 \hspace{0.5cm} \Rightarrow 16g+2f+c = -65$ … … … … … iii)

Substituting $\displaystyle (1,3)$ in i) we get

$\displaystyle 1 + 9 + 2g + 6f + c = 0 \hspace{0.5cm} \Rightarrow 2g+6f+c = -10$ … … … … … iv)

Solving ii) , iii) and iv) simultaneously, we get

$\displaystyle g = \frac{-29}{6} , \hspace{0.5cm} f = \frac{-19}{6} , \hspace{0.5cm} c= \frac{56}{3}$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 - \frac{29}{3} x - \frac{19}{3} y + \frac{56}{3} = 0$

$\displaystyle \Rightarrow 3(x^2+y^2) - 39x-19y+56=0$

(ii) The circle passes through the points $\displaystyle (1,2), (3, - 4)$ ,and $\displaystyle (5, - 6)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle (1,2)$ in i) we get

$\displaystyle 1 + 4 + 2g + 4f + c = 0 \hspace{0.5cm} \Rightarrow 2g+4f+c = -5$ … … … … … ii)

Substituting $\displaystyle (3,-4)$ in i) we get

$\displaystyle 9 + 16 + 6g -8f + c = 0 \hspace{0.5cm} \Rightarrow 6g-8f+c = -25$ … … … … … iii)

Substituting $\displaystyle (5, -6)$ in i) we get

$\displaystyle 25 +36 + 10g -12f + c = 0 \hspace{0.5cm} \Rightarrow 10g-12f+c = -61$ … … … … … iv)

Solving ii) , iii) and iv) simultaneously, we get

$\displaystyle g = -11, \hspace{0.5cm} f = -2, \hspace{0.5cm} c= 25$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 - 22 x --4 y + 25 = 0$

(iii) The circle passes through the points $\displaystyle (5, - 8), (- 2,9)$ and $\displaystyle (2,1)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle (5,-8)$ in i) we get

$\displaystyle 25 + 64 + 10g -16f + c = 0 \hspace{0.5cm} \Rightarrow 10g-16f+c = -91$ … … … … … ii)

Substituting $\displaystyle (-2, 9)$ in i) we get

$\displaystyle 4 + 81 -4g +18f + c = 0 \hspace{0.5cm} \Rightarrow -4g+18f+c = -85$ … … … … … iii)

Substituting $\displaystyle (2, 1)$ in i) we get

$\displaystyle 4 +1 + 4g +2f + c = 0 \hspace{0.5cm} \Rightarrow 4g+2f+c = -5$ … … … … … iv)

Solving ii) , iii) and iv) simultaneously, we get

$\displaystyle g = 58, \hspace{0.5cm} f = 24, \hspace{0.5cm} c= -285$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 +116 x +48 y -285 = 0$

(iv) The circle passes through the points $\displaystyle (0, 0), (-2, 1)$ and $\displaystyle (- 3,2)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle (0,0)$ in i) we get

$\displaystyle 0 + 0 + 0 + 0 + c = 0 \hspace{0.5cm} \Rightarrow c = 0$ … … … … … ii)

Substituting $\displaystyle (-2,1)$ in i) we get

$\displaystyle 4 + 1 -4g + 2f + c = 0 \hspace{0.5cm} \Rightarrow -4g+2f = 0$ … … … … … iii)

Substituting $\displaystyle (-3, 2)$ in i) we get

$\displaystyle 9 + 4 -6g + 4f + c = 0 \hspace{0.5cm} \Rightarrow -6g+4f = -13$ … … … … … iv)

Solving ii) , iii) and iv) simultaneously, we get

$\displaystyle g = \frac{-3}{2} , \hspace{0.5cm} f = \frac{-11}{2} , \hspace{0.5cm} c= 0$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 - 3 x -11 y = 0$

$\displaystyle \\$

Question 3: Find the equation of the circle which passes through $\displaystyle (3,1), (-2, 0)$ and has its center on the line $\displaystyle 2x-y=3$ .

Answer:

The circle passes through $\displaystyle (3,1), (-2, 0)$ and has its center on the line $\displaystyle 2x-y=3$ .

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle (3, -2)$ in i) we get

$\displaystyle 9+4+6g-4f+c = 0 \hspace{0.5cm} \Rightarrow 6g-4f+c = -13$ … … … … … ii)

Substituting $\displaystyle (-2,0)$ in i) we get

$\displaystyle 4 + 0 -4g + 0 + c = 0 \hspace{0.5cm} \Rightarrow -4g+c = -4$ … … … … … iii)

Since center $\displaystyle (-g, -f)$ lie on $\displaystyle 2x-y=3$ , we get

$\displaystyle -2g+f = 3$ … … … … … iv)

Solving ii) , iii) and iv) simultaneously, we get

$\displaystyle g = \frac{3}{2} , \hspace{0.5cm} f = 6 , \hspace{0.5cm} c= 2$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 + 3 x +12 y +2 = 0$

$\displaystyle \\$

Question 4: Find the equation of the circle which passes through the points $\displaystyle (3,7), (5,5)$ and has its center on the line $\displaystyle x - 4y =1$ .

Answer:

The circle passes through $\displaystyle (3,7), (5,5)$ and has its center on the line $\displaystyle x - 4y =1$ .

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle (3, 7)$ in i) we get

$\displaystyle 9+49+6g+14f+c = 0 \hspace{0.5cm} \Rightarrow 6g+14f+c = -58$ … … … … … ii)

Substituting $\displaystyle (5,5)$ in i) we get

$\displaystyle 25 + 25 +10g + 10f + c = 0 \hspace{0.5cm} \Rightarrow 10g + 10f + c = -50$ … … … … … iii)

Since center $\displaystyle (-g, -f)$ lie on $\displaystyle 2x-y=3$ , we get

$\displaystyle -g+4f = 1$ … … … … … iv)

Solving ii) , iii) and iv) simultaneously, we get

$\displaystyle g = 3 , \hspace{0.5cm} f = 1 , \hspace{0.5cm} c= -90$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 + 6 x +2 y -90 = 0$

$\displaystyle \\$

Question 5: Show that the points $\displaystyle (3, - 2), (1, 0), (-1, - 2)$ and $\displaystyle (1, - 4)$ are concyclic.

Answer:

Consider $\displaystyle P(3, - 2), Q(1, 0), R(-1, - 2)$ and $\displaystyle S(1, - 4)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle P(3, -2)$ in i) we get

$\displaystyle 9 + 4 + 6g -4f + c = 0 \hspace{0.5cm} \Rightarrow 6g-4f+c = -13$ … … … … … ii)

Substituting $\displaystyle Q(1, 0)$ in i) we get

$\displaystyle 1 + 0 +2g + 0 + c = 0 \hspace{0.5cm} \Rightarrow 2g+ c = -1$ … … … … … iii)

Substituting $\displaystyle R(-1, 2)$ in i) we get

$\displaystyle 1 + 4 -2g + 4f + c = 0 \hspace{0.5cm} \Rightarrow -2g+4f +c = -5$ … … … … … iv)

Solving ii) , iii) and iv) simultaneously, we get

$\displaystyle g = -1 , \hspace{0.5cm} f = 2 , \hspace{0.5cm} c= 1$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 - 2 x +4 y+1 = 0$ … … … … … v)

Now we check if $\displaystyle S( 1, -4)$ satisfies equation v)

$\displaystyle 1^2 + ( -4)^2 - 2 ( 1) + 4 ( -4) +1 = 0$ . Therefore S lies on the circle.

Therefore, $\displaystyle P(3, - 2), Q(1, 0), R(-1, - 2)$ and $\displaystyle S(1, - 4)$ are con-cyclic.

$\displaystyle \\$

Question 6: Show that the points $\displaystyle (5,5), (6, 4), (-2,4)$ and $\displaystyle (7,1)$ all lie on a circle, and find its equation, center and radius.

Answer:

Consider $\displaystyle P(5,5), Q(6, 4), R(-2,4)$ and $\displaystyle S(7,1)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle P(5,5)$ in i) we get

$\displaystyle 25 + 25 + 10g +10f + c = 0 \hspace{0.5cm} \Rightarrow 10g +10f + c = -50$ … … … … … ii)

Substituting $\displaystyle Q(6,4)$ in i) we get

$\displaystyle 36 + 16 +12g + 8f + c = 0 \hspace{0.5cm} \Rightarrow 12g + 8f + c = -52$ … … … … … iii)

Substituting $\displaystyle R(-2,4)$ in i) we get

$\displaystyle 4 + 16 -4g + 8f + c = 0 \hspace{0.5cm} \Rightarrow -4g + 8f + c = -20$ … … … … … iv)

Solving ii) , iii) and iv) simultaneously, we get

$\displaystyle g = -2 , \hspace{0.5cm} f = -1 , \hspace{0.5cm} c= -20$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 - 4 x -2 y -20 = 0$ … … … … … v)

Now we check if $\displaystyle S( 7, 1)$ satisfies equation v)

$\displaystyle 7^2 + 1^2 - 4 ( 7) -2 ( 1) -20 = 0$ . Therefore S lies on the circle.

Therefore, $\displaystyle P(5,5), Q(6, 4), R(-2,4)$ and $\displaystyle S(7,1)$ are con-cyclic.

Therefore center $\displaystyle = ( -g, -f) = ( 2, 1)$

Radius $\displaystyle = \sqrt{g^2+f^2-c} = \sqrt{4+1+20} = \sqrt{25} = 5$ units.

$\displaystyle \\$

Question 7: Find the equation of the circle which circumscribes the triangle formed by the lines:

i) $\displaystyle x+y+3=0 , x-y+1=0$ and $\displaystyle x = 3$

ii) $\displaystyle 2x+y-3=0, x+y-1=0$ and $\displaystyle 3x+2y - 5 = 0$

iii) $\displaystyle x+y = 2, 3x-4y=6$ and $\displaystyle x - y = 0$

iv) $\displaystyle y = x + 2 , 3y = 4x$ and $\displaystyle 2y = 3x$

Answer:

i) Given equations:

$\displaystyle x+y+3=0 \hspace{0.5cm} \Rightarrow x+y = - 3$ … … … … … i)

$\displaystyle x-y+1=0 \hspace{0.5cm} \Rightarrow x-y = - 1$ … … … … … ii)

$\displaystyle x = 3$ … … … … … iii)

Solving i) , ii) and iii) we get the vertices $\displaystyle A( -2, 1), B( 3, 4), C( 3, -6)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … iv)

Substituting $\displaystyle A( -2, 1)$ in iv) we get

$\displaystyle 4 + 1 -4g +2f + c = 0 \hspace{0.5cm} \Rightarrow -4g +2f + c = -5$ … … … … … v)

Substituting $\displaystyle B( 3, 4)$ in iv) we get

$\displaystyle 9 + 16 +6g + 8f + c = 0 \hspace{0.5cm} \Rightarrow 6g + 8f + c = -25$ … … … … … vi)

Substituting $\displaystyle C( 3, -6)$ in iv) we get

$\displaystyle 9 + 36 +6g -12f + c = 0 \hspace{0.5cm} \Rightarrow 6g -12f + c = -45$ … … … … … vii)

Solving v) , vi) and vii) simultaneously, we get

$\displaystyle g = -3 , \hspace{0.5cm} f = 1 , \hspace{0.5cm} c= -15$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 - 6 x +2 y -15 = 0$

ii) Given equations:

$\displaystyle 2x+y-3=0 \hspace{0.5cm} \Rightarrow 2x+y = 3$ … … … … … i)

$\displaystyle x+y-1=0 \hspace{0.5cm} \Rightarrow x+y = 1$ … … … … … ii)

$\displaystyle 3x+2y - 5 = 0 \hspace{0.5cm} \Rightarrow 3x+2y = 5$ … … … … … iii)

Solving i) , ii) and iii) we get the vertices $\displaystyle A( 2, -1), B( 3, -2), C( 1,1)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … iv)

Substituting $\displaystyle A( 2, -1)$ in iv) we get

$\displaystyle 4 + 1 +4g -2f + c = 0 \hspace{0.5cm} \Rightarrow 4g -2f + c = -5$ … … … … … v)

Substituting $\displaystyle B( 3, -2)$ in iv) we get

$\displaystyle 9 + 4 +6g -4f + c = 0 \hspace{0.5cm} \Rightarrow 6g -4f + c = -13$ … … … … … vi)

Substituting $\displaystyle C( 1,1)$ in iv) we get

$\displaystyle 1 + 1 +2g +2f + c = 0 \hspace{0.5cm} \Rightarrow 2g +2f + c = -2$ … … … … … vii)

Solving v) , vi) and vii) simultaneously, we get

$\displaystyle g = \frac{-13}{2} , \hspace{0.5cm} f = \frac{-5}{2} , \hspace{0.5cm} c= 16$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 - 13 x -5 y +16 = 0$

iii) Given equations:

$\displaystyle x+y = 2$ … … … … … i)

$\displaystyle 3x-4y=6$ … … … … … ii)

$\displaystyle x - y = 0$ … … … … … iii)

Solving i) , ii) and iii) we get the vertices $\displaystyle A(2,0), B( -6,-6), C( 0,0 )$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … iv)

Substituting $\displaystyle A( 2,0 )$ in iv) we get

$\displaystyle 4 + 0 +4g +0 + c = 0 \hspace{0.5cm} \Rightarrow 4g + c = -4$ … … … … … v)

Substituting $\displaystyle B( -6,-6 )$ in iv) we get

$\displaystyle 36 + 36 -12 g -12 f + c = 0 \hspace{0.5cm} \Rightarrow -12g -12f + c = -72$ … … … … … vi)

Substituting $\displaystyle C( 1,1)$ in iv) we get

$\displaystyle 1 + 1 +2g +2f + c = 0 \hspace{0.5cm} \Rightarrow 2g +2f + c = -2$ … … … … … vii)

Solving v) , vi) and vii) simultaneously, we get

$\displaystyle g = 2 , \hspace{0.5cm} f = 3 , \hspace{0.5cm} c= -12$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 +4 x +6 y -12 = 0$

iv) Given equations:

$\displaystyle y = x + 2 \hspace{0.5cm} \Rightarrow -x+y = 2$ … … … … … i)

$\displaystyle 3y = 4x \hspace{0.5cm} \Rightarrow -4x+3y=0$ … … … … … ii)

$\displaystyle 2y = 3x \hspace{0.5cm} \Rightarrow -3x+2y=0$ … … … … … iii)

Solving i) , ii) and iii) we get the vertices $\displaystyle A(4, 6), B( 6, 8 ), C( 0, 0 )$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … iv)

Substituting $\displaystyle A( 4, 6 )$ in iv) we get

$\displaystyle 16 + 36 +8 g +2f + c = 0 \hspace{0.5cm} \Rightarrow 8 g +2f + c = -54$ … … … … … v)

Substituting $\displaystyle B( 6, 8 )$ in iv) we get

$\displaystyle 36 + 64 +12g + 16f + c = 0 \hspace{0.5cm} \Rightarrow 12g + 16f + c = -100$ … … vi)

Substituting $\displaystyle C( 3, -6)$ in iv) we get

$\displaystyle 0 + 0 +0 +0 + c = 0 \hspace{0.5cm} \Rightarrow c = 0$ … … … … … vii)

Solving v) , vi) and vii) simultaneously, we get

$\displaystyle g = -23 , \hspace{0.5cm} f = 21 , \hspace{0.5cm} c= 0$

Therefore the equation of the required circle is:

$\displaystyle x^2 + y^2 - 46 x +24 y = 0$

$\displaystyle \\$

Question 8: Prove that the centers of the three circles $\displaystyle x^2 + y^2 - 4x - 6y -12=0$ , $\displaystyle x^2 + y^2 + 2x + 4y -10=0$ and $\displaystyle x^2 + y^2 - 10x- 16y -1 =0$ are collinear.

Answer:

Given equations of the circle are:

$\displaystyle x^2 + y^2 - 4x - 6y -12=0 \hspace{0.5cm} \Rightarrow \text{ Center } = ( 2, 3)$

$\displaystyle x^2 + y^2 + 2x + 4y -10=0 \hspace{0.5cm} \Rightarrow \text{ Center } = ( -1,-2)$

$\displaystyle x^2 + y^2 - 10x- 16y -1 =0 \hspace{0.5cm} \Rightarrow \text{ Center }= ( 5,8)$

Area of the circle formed between the three centers

Area $\displaystyle = \frac{1}{2} | 2( -2-8) -1( 8 -3) + 5 ( 3+2) | = \frac{1}{2} | -20 - 5 + 25 | = 0$

Therefore the three centers are collinear.

$\displaystyle \\$

Question 9: Prove that the radii of the circles $\displaystyle x^2 + y^2 =1$ , $\displaystyle x^2 + y^2 -2x-6y-6=0$ and $\displaystyle x^2 + y^2 -4x-12y -9 = 0$ are in AP.

Answer:

Given equations of the circle are:

$\displaystyle x^2 + y^2 =1 \hspace{0.5cm} \Rightarrow \text{ Radius } = r_1 = 1$

$\displaystyle x^2 + y^2 -2x-6y-6=0$

Radius $\displaystyle = r_2 = \sqrt{ g^2 + f^2 - c } = \sqrt{ (-1)^2 + ( -3)^2 - ( -6) } = \sqrt{16} = 4$

$\displaystyle x^2 + y^2 -4x-12y -9 = 0$

Radius $\displaystyle = r_3 = \sqrt{ g^2 + f^2 - c } = \sqrt{ (-2)^2 + ( -6)^2 - ( -9) } = \sqrt{49} = 7$

We see that $\displaystyle 2r_2 = r_1 + r_3$

Therefore $\displaystyle r_1, r_2 , r_3$ are in A.P.

$\displaystyle \\$

Question 10: Find the equation of the circle which passes through the origin and cuts off chords of lengths $\displaystyle 4$ and $\displaystyle 6$ on the positive side of the x-axis and y-axis respectively.

Answer:

Please refer to adjoining figure.

Center $\displaystyle ( C) =$ Mid point of $\displaystyle AB = ( 2, 3)$

Therefore the equation of the circle is:

$\displaystyle x^2 + y^2 + 2(-2)x + 2 ( -3) y = 0$ $\displaystyle \Rightarrow x^2 + y^2 - 4x - 6y = 0$

$\displaystyle \\$

Question 11: Find the equation of the circle concentric with the circle $\displaystyle x^2 + y^2 - 6x + 12y + 15 = 0$ and double of its area.

Answer:

Given equation: $\displaystyle x^2 + y^2 - 6x + 12y + 15 = 0$

Therefore center $\displaystyle = ( -g, -f) = ( 3, -6)$

Radius $\displaystyle = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+6^2 - 15} = \sqrt{30}$

The center remains the same but the area of the new circle is double that of the original circle. Therefore, the radius of the new circle:

$\displaystyle 2 \times [ \pi ( \sqrt{30})^2 ] = \pi r^2 \hspace{0.5cm} \Rightarrow r = 2\sqrt{15}$

Therefore the equation of the required circle is

$\displaystyle (x-3)^2 + [ y - ( -6) ] ^2 = (2\sqrt{15})^2$

$\displaystyle \Rightarrow x^2 + 9 - 6x + y^2 + 36 + 12y = 60$

$\displaystyle \Rightarrow x^2 + y^2 - 6x + 12y - 15 = 0$

$\displaystyle \\$

Question 12: Find the equation to the circle which passes through the points $\displaystyle (1,1) (2,2)$ and whose radius is $\displaystyle 1$ . Show that there are two such circles.

Answer:

We know that the general equation of the circle is $\displaystyle ( x-h)^2 + ( y - k)^2 = r^2$ . Since the circle’s radius is $\displaystyle 1$ , we have

$\displaystyle ( x-h)^2 + ( y - k)^2 = 1$ … … … … … i)

The circle passes through $\displaystyle (1,1)$ and $\displaystyle ( 2, 2)$ . Therefore,

$\displaystyle ( 1- h)^2 + ( 1-k)^2 = 1$ … … … … … ii)

$\displaystyle ( 2- h)^2 + ( 2-k)^2 = 1$ … … … … … iii)

$\displaystyle \therefore ( 1- h)^2 + ( 1-k)^2 = ( 2- h)^2 + ( 2-k)^2$

$\displaystyle \Rightarrow 1 + h^2 - 2h + 1 + k^2 - 2k = 4 + h^2 - 4h + 4 + k^2 - 4k$

$\displaystyle \Rightarrow 2 - 2h - 2k = 8 - 4h - 4k$

$\displaystyle \Rightarrow 2h + 2k = 6$

$\displaystyle \Rightarrow h + k = 3$

$\displaystyle \Rightarrow h = 3-k$ … … … … … iv)

Substituting iv) in ii) we get

$\displaystyle [ 1 - ( 3-k)]^2 + ( 1- k)^2 = 1$

$\displaystyle \Rightarrow ( -2 + k) ^2 + ( 1-k)^2 = 1$

$\displaystyle \Rightarrow 4 + k^2 - 4k + 1 + k^2 - 2k = 1$

$\displaystyle \Rightarrow 2k^2 - 6k + 4=0$

$\displaystyle \Rightarrow k^2 - 3k + 2 = 0$

$\displaystyle \Rightarrow ( k-1)(k-2) = 0$

$\displaystyle \Rightarrow k = 1$ or $\displaystyle \Rightarrow k = 2$

When $\displaystyle k = 2, \hspace{0.5cm} \Rightarrow h = 3-2 = 1$

When $\displaystyle k = 1, \hspace{0.5cm} \Rightarrow h = 3-1 = 2$

Therefore the equation of the circle are:

$\displaystyle ( x-2)^2 + ( y - 1)^2 = 1 \hspace{0.5cm} \Rightarrow x^2 + y^2 - 4x - 2y + 4 = 0$

$\displaystyle ( x-1)^2 + ( y - 2)^2 = 1 \hspace{0.5cm} \Rightarrow x^2 + y^2 - 2x - 4y + 4 = 0$

$\displaystyle \\$

Question 13: Find the equation of the circle concentric with $\displaystyle x^2 + y^2 - 4x - 6y- 3 = 0$ and which touches the y-axis.

Answer:

Given the circle is $\displaystyle x^2 + y^2 - 4x - 6y- 3 = 0$ .

Therefore the center $\displaystyle = ( - g, -f) = ( 2, 3)$

The required circle is concentric with the given center, therefore center of the required circle is also $\displaystyle ( 2, 3)$

This circle also touches y-axis , therefore the radius $\displaystyle = 2$

Therefore the equation of the circle is:

$\displaystyle (x-2)^2 + ( y - 3)^2 = 2^2 \hspace{0.5cm} \Rightarrow x^2 + y^2 - 4x - 6y + 9 = 0$

$\displaystyle \\$

Question 14: If a circle passes through the points $\displaystyle (0,0), (a,0), (0,b)$ , then find the coordinates of its center.

Answer:

The circle passes through the points $\displaystyle (0,0), (a,0), (0,b)$

The equation of circle is $\displaystyle x^2 + y^2 + 2gx + 2fy + c =0$ … … … … … i)

Substituting $\displaystyle (0,0 )$ in i) we get

$\displaystyle 0 + 0 + 0 + 0+ c = 0 \hspace{0.5cm} \Rightarrow c = 0$ … … … … … ii)

Substituting $\displaystyle (a,0)$ in i) we get

$\displaystyle a^2 + 2ag = 0 \hspace{0.5cm} \Rightarrow a( a+2g) = 0 \hspace{0.5cm} \Rightarrow g = \frac{-a}{2}$ … … … … … iii)

Substituting $\displaystyle (0, b)$ in i) we get

$\displaystyle b^2 + 2bf = 0 \hspace{0.5cm} \Rightarrow b(b+2f) = 0 \hspace{0.5cm} \Rightarrow f = \frac{-b}{2}$ … … … … … iv)

Therefore center $\displaystyle = ( -g, -f) = ( \frac{a}{2} , \frac{b}{2} )$

$\displaystyle \\$

Question 15: Find the equation of the circle which passes through the points $\displaystyle (2, 3)$ and $\displaystyle (4,5)$ and the center lies on the straight line $\displaystyle y-4x+3=0$