Question

If the circle $C_1:x^2+y^2=16$ intersects another circle $C_2$ of radius $5$ in such a manner that common chord is of maximum length and has a slope equal to $\frac{3}{4}$, then the absolute sum of coordinates of the centre $C_2$ is

Answer 1

We have $C_1:x^2+y^2=16$
Centre $O_1(0,0)$, radius $=4$.
$C_2$ is another circle with radius $5$, let its centre $O_2$ be $(h,k)$.

Now, the common chord of circles $C_1$ and $C_2$ is of maximum length when chord is diameter of the smaller circle $C_1$, and then it passes through centre $O_1$ of circle $C_1$.
Given that slope of this chord is $\frac{3}{4}$.
$\therefore$ Equation of $AB$ is,
$y=\frac{3}{4}x$
$\Rightarrow 3x-4y=0\dots \left(1\right)$

In right $\Delta A{O}_1{O}_2$,
$O_1{O}_2=\sqrt{5^2-4^2}=3$
Also $O_1{O}_2={\perp }^r$ distance from $(h,k)$ to $\left(1\right)$
$\Rightarrow 3=\left|\frac{3h-4k}{\sqrt{3^2+4^2}}\right|$
$\Rightarrow \pm 3=\frac{3h-4k}{5}$
$\Rightarrow 3h-4k\pm 15=0\dots \left(2\right)$

Again, $AB\perp O_1{O}_2$
$\Rightarrow m_{AB}\times m_{O_1{O}_2}=-1$
$\Rightarrow \frac{3}{4}\times \frac{k}{h}=-1$
$\Rightarrow 4h+3k=0\dots \left(3\right)$
Solving $3h-4k+15=0$ and $4h+3k=0,$
we get $h=-\frac{9}{5},k=\frac{12}{5}$
Again solving $3h-4k-15=0$ and $4h+3k=0,$
we get $h=\frac{9}{5},k=-\frac{12}{5}$
Thus the required centre is $(\frac{-9}{5},\frac{12}{5})$ or $(\frac{9}{5},\frac{-12}{5})$
$\therefore$ Absolute sum of coordinates of centre $C_2=\frac{3}{5}$