Question

Find the equation of family of circles passing through the point of intersection of the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-10x-12y+40=0$ and whose radius is 4.

• A. $x^2+y^2+2y+15=0$
• B. $2x^2+2y^2-18x-22y+69=0$
• C. $x^2+y^2-2y-15=0$
• D. $x^2+y^2-18x-22y+69=0$

Answer 1

Answer: (B), (C)

$x^2+y^2-2y-15=0$

Family of circles through intersection of circles is given by

$S_1+\lambda S_2=0$

$(x^2+y^2-2x-4y-4)+\lambda (x^2+y^2-10x-12y+40)=0$

$(1+\lambda )x^2+(1+\lambda )y^2-2(1+5\lambda )x-2(2+6\lambda )y+40\lambda -4=0$

$x^2+y^2-\frac{2(1+5\lambda )}{1+\lambda }x+\frac{2(2+6\lambda )}{1+\lambda }y+\frac{40\lambda -4}{1+\lambda }=0$ - - - - - - (1)

center of the circle (-g,-f)

$(\frac{1+5\lambda }{1+\lambda },\frac{2+6\lambda }{1+\lambda })$

Given radius of the required circles is =4

$\sqrt{g^2+f^2-c}=4$

${\left(\frac{1+5\lambda }{1+\lambda }\right)}^2+{\left(\frac{2+6\lambda }{1+\lambda }\right)}^2-\frac{40\lambda -4}{1+\lambda }=16$

$16(1+{\lambda }^2+2\lambda )=1+10\lambda +25{\lambda }^2+4+24\lambda +36{\lambda }^2-40{\lambda }^2+4+4\lambda$

$16+332\lambda +16{\lambda }^2=21{\lambda }^2-2\lambda +9$

$5{\lambda }^2-34\lambda -7=0$

$(\lambda -7)(5\lambda +1)=0$

$\lambda =7,\frac{-1}{5}$

Substituting these values of $\lambda$ in equation (1)

The required circles are

when $\lambda =7$

$x^2+y^2-\frac{2(1+5\times 7)}{1+7}-\frac{2(2+6\times 7)}{1+7}+\frac{40\times 7-4}{1+7}=0$

$X^2+y^2-9x-11y+\frac{69}{2}=0$

When $\lambda =\frac{-1}{5}$

$x^2+y^2-2\frac{(1+5\left(\frac{-1}{5}\right))}{1-\frac{1}{5}}-2\frac{(2+6\left(\frac{-1}{5}\right))}{1-\frac{1}{5}}+\frac{40\left(\frac{-1}{5}\right)-4}{1-\frac{1}{5}}$

$x^2+y^2-0-2y-15=0$

$x^2+y^2-2y-15=0$

Requiredequations of family of circle are $2x^2+2y^2-18x-22y+69=0$ and $x^2+y^2-2y-15=0$