Question

Lines $5x+12y-10=0$ and $5x-12y-40=0$ touch circle $C_1$ of diameter $6$. If the centre of $C_1$ lies in the first quadrant, find the equation of the circle $C_2$ which is concentric with $C_1$ and cuts circle $C_2$ which is concentric with $C_1$ and cuts intercept of length $8$ on these lines.

Answer 1

If $C$ be the centre of the circle $C_1$ then it lies on the bisectors of the given lines which are
$\frac{5x+12y-10}{13}=\pm \frac{5x-12y-40}{13}$
These give $x=5$ and $y=-5/4$. Since centre lies in $1st$ quadrant therefore $y=-5/4$ is ruled out. Let the centre be $(5,k)$ then its perpendicular distance from each of lines will be $3$,
i.e. $p=r$.
$\therefore \frac{25+12kl-10}{13}=3$ and $\frac{25-12k-40}{13}=3$
$\therefore k=2$ or $k=-\frac{54}{12}=-\frac{9}{2}$
The value $-9/2$ of $k$ is ruled out as centre $C$ lies in $1st$ quadrant $\therefore C$ is $(5,2)$.
Now the circle $C_2$ cuts off intercepts $8$ from these lines and if its radius be $r$, then
$r^2=3^2+4^2=25$ or $r=5$
$\therefore$ Circle $C_2$ is $(x-5{)}^2+(y-2{)}^2=5^2$
or $x^2+y^2-10x-4y+4=0$.