Question

If a circle $C_{1}: x^{2}+y^{2}=16$ intersects another circle $C_{2}$ with radius 5 such that the common chord is of maximum length and has a slope equal to $\frac{3}{4}$ then the centre of the circle $C_{2}$ is

• A. $\left(-\frac{9}{5}, \frac{12}{5}\right)$
• B. $\left(\frac{9}{5}, \frac{12}{5}\right)$
• C. $\left(-\frac{5}{9}, \frac{6}{5}\right)$
• D. $\left(\frac{7}{5},-\frac{12}{5}\right)$

Answer 1

Answer: (A)

Given circle, $C_{1}: x^{2}+y^{2}=16$
radius $=4$ units, centre $(0,0)$ The maximum length of chord $=$ the diameter of circle $c_{1}=8$ units. The equation of chord passing through $(0,0)$ and slope $\frac{3}{4}$ is
$y=\frac{3}{4} \,x$
$ \Rightarrow \,3 x-4 y=0$
The centre of circle $C_{2}$ must be on the line perpendicular to the chord.
So, the coordinate of the centre of circle can be written as $(3 a,-4 a)$ in $\Delta A O_{1} O_{2}$

$\left(O_{1} \,O_{2}\right)^{2}=5^{2}-4^{2} $
$ O_{1} O_{2}=\sqrt{9}$
$O_{1} \,O_{2}=3$
$O_{1} \,O_{2} \perp \,A B$, so distance of $(3 a,-4 a)$ from $3 x-4 y$ is 3 units
$\left|\frac{+3(3 a)-4(-4 a)}{\sqrt{3^{2}+(-4)^{2}}}\right| =3$
$ \left|\frac{-25 a}{5}\right|=3$
$ \Rightarrow \, a=\pm \frac{3}{5}$
if $ a=\frac{3}{5}$
Then, $O_{2}\left(\frac{9}{5},-4\left(\frac{3}{5}\right)\right)$
$O_{2}\left(\frac{9}{5},-\frac{12}{5}\right)$
and if $a=\frac{-3}{5}$
Then, coordinate of $O_{2}\left(3\left(\frac{-3}{5}\right),-4\left(\frac{-3}{5}\right)\right)$
$O_{2}\left(\frac{-9}{5}, \frac{12}{5}\right)$