Question

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

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

Answer 1

The correct option is B $(-\frac{9}{5},\frac{12}{5})$

The common chord is longest when it is a diameter of the smaller circle.
The radius of $C_2$ is $5$(given)
The distance between the two centres $O_1{O}_2$ is $3$ units.
$\therefore$The centre of $C_2$ lies on the line with slope $-\frac{4}{3}$
$\therefore O_2=(3\cos \theta ,3\sin \theta )$ where $\tan \theta =-\frac{4}{3}$
$\sin \theta =-\frac{4}{5},\cos \theta =\frac{3}{5}$
$\Rightarrow O_2=(\frac{9}{5},-\frac{12}{5})$
$\sin \theta =\frac{4}{5},\cos \theta =-\frac{3}{5}$
$\Rightarrow O_2=(-\frac{9}{5},\frac{12}{5})$