Question

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

Answer 1

Given, $C_1:x^2+y^2=16$
and let $C_2:(x-h{)}^2+(y-k{)}^2=25$
$\therefore$ Equation of common chords is $S_1-S_2=0$
$\therefore 2hx+2ky=(h^2+k^2-9)$
$\therefore$ Its slope = $-\frac{h}{k}=\frac{3}{4}$
If $p$ be the length of perpendicular on it from the centre
$(0,0)$ of $C_1$ of radius $4$, then
$p=\frac{h^2+k^2-9}{\sqrt{4h^2+4k^2}}$
Also, the length of the chord is
$2\sqrt{r^2-p^2}=2\sqrt{4^2-p^2}$
The chord will be of maximum length, if $\varphi =0$ or
$h^2+k^2-9=0\Rightarrow h^2+\frac{16}{9}{h}^2=9\Rightarrow h=\pm \frac{9}{5}$
$\therefore k=\mp \frac{12}{5}$
Hence, centres are $(\frac{9}{5},\frac{-12}{5})$ and $(-\frac{9}{5},\frac{12}{5})$