Question

Find the equations of the circles passing through the points 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 $4$

Answer 1

Any circle through the intersection of given circles is $S_1+\lambda S_2=0$
or $(x^2+y^2-2x-4y-4)+\lambda (x^2+y^2-10x-12y+40)=0$
or $(x^2+y^2)-2\frac{(1+5\lambda )}{1+\lambda }x-2\frac{(2+6\lambda )}{1+\lambda }y+\frac{40\lambda -4}{1+\lambda }=\mathrm{0...........}\left(i\right)$
$r=\sqrt{g^2+f^2-c}=4$ given
$\therefore 16=\frac{(1+5\lambda {)}^2}{(1+\lambda {)}^2}+\frac{(2+6\lambda {)}^2}{(1+\lambda {)}^2}-\frac{40\lambda -4}{1+\lambda }$
$16(1+2\lambda +{\lambda }^2)=1+10\lambda +25{\lambda }^2+4+24\lambda +36{\lambda }^2-40{\lambda }^2-40\lambda +4+4\lambda$
or $16+32\lambda +16{\lambda }^2=21{\lambda }^2-2\lambda +9$
or $5{\lambda }^2-34\lambda -7=0$
$\therefore (\lambda -7)(5\lambda +1)=0\therefore \lambda =7,-1/5$
Putting the values of $\lambda$ in (i) the required circles are $2x^2+2y^2-18x-22y+69=0$
and $x^2+y^2-2y-15=0$