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
$y=\frac{3}{4}x$
$\Rightarrow 3x-4y=0$
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 $(3a,-4a)$ in $\Delta A{O}_1{O}_2$

${\left(O_1{O}_2\right)}^2=5^2-4^2$
$O_1{O}_2=\sqrt{9}$
$O_1{O}_2=3$
$O_1{O}_2\perp AB$, so distance of $(3a,-4a)$ from $3x-4y$ is 3 units
$\left|\frac{+3(3a)-4(-4a)}{\sqrt{3^2+(-4{)}^2}}\right|=3$
$\left|\frac{-25a}{5}\right|=3$
$\Rightarrow a=\pm \frac{3}{5}$
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 $O_2\left(3\left(\frac{-3}{5}\right),-4\left(\frac{-3}{5}\right)\right)$
$O_2\left(\frac{-9}{5},\frac{12}{5}\right)$