Question

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

• A. $(\pm \frac{9}{5},\mp \frac{12}{5})$
• B. $(\pm \frac{9}{5},\pm \frac{12}{5})$
• C. $(\pm \frac{3}{2},\mp 2)$
• D. $(\pm \frac{3}{2},\pm 2)$

Answer 1

The correct option is A $(\pm \frac{9}{5},\mp \frac{12}{5})$

Let the coordinates of the centre of circle $C_2$ is $(h,k)$.
The length of the chord is maximum when it is diameter of smaller circle.



Here, $AB$ is a diameter of the circle $C_1.$
$C_1{C}_2=\sqrt{h^2+k^2}=3\Rightarrow h^2+k^2=9.....\left(i\right)$
Slope of $AB=\frac{3}{4}$ and $C_1{C}_2\perp AB$
$\Rightarrow \frac{k}{h}\times \frac{3}{4}=-1.....\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$(h,k)=(\pm \frac{9}{5},\mp \frac{12}{5})$