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 $\phi= 0$ 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 $ \Big( \frac{9}{5},\frac{-12}{5}\Big) $ and $ \Big( -\frac{9}{5},\frac{12}{5}\Big) $