Question

Find the equation of a circle of radius $5$ which lies within the circle $x^2+y^2-14x-10y-26=o$ and touches the given circle at the point .$(-1,3)$

Answer 1

$x^2+y^2-14x-10y-26=0-----\left(1\right)$

Comparing with standard equation $x^2+y^2+2gx+2fy+c=0$

$g=-7$ and $f=-5,c=-26$

Centre of $\left(1\right)=(-g,-f)=(+7,+5)$

Radius of $\left(1\right)=\sqrt{g^2+f^2-c}=\sqrt{47+25+26}=\sqrt{100}=10$

Radius of circle where equation is to be found $=5$

Two circles touch at point $(-1,3)$ and the circle lies inside of circle $\left(1\right)$

So, centre of internal circle is mid-point of $(7,5)$ and $(-1,3):(\frac{7-1}{2},\frac{5+3}{2})=(g,t)=(3,4)$

$\therefore$ Required radius $\Rightarrow 5=\sqrt{9+16-c}$

Or, $25-c=25\Rightarrow c=-1$

$\therefore$Required equation of internal circle:

$x^2+y^2+6x+8y-1=0$