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Q.
The radical axis of the pairs of circles $x^{2}+y^{2}=144$ and $x^{2}+y^{2}-15 x+11 y=0$, is
Conic Sections
Solution:
The equations of the circle are
$ S _{1}= x ^{2}+ y ^{2}-144=0$
$S _{2}= x ^{2}+ y ^{2}-15 x +11 y =0$
$\therefore $ The equation of the radical axis is
$S _{1}- S _{2}=0$
$\Rightarrow \left( x ^{2}+ y ^{2}-144\right)-\left( x ^{2}+ y ^{2}-15 x +11 y \right)=0 $
or $ 15 x - 11 y -144=0 $
So, the required equation is $15 x-11 y-144=0$.