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 3/4, the coordinates of the centre of $C_2$ are

• 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

Answer: (A), (B)

The correct options are
A $(\frac{9}{5},\frac{-12}{5})$
B $(\frac{-9}{5},\frac{12}{5})$

When two circles intersect, the common chord of maximum length will be the diameter of the smaller circle.

Let $O_2(\alpha ,\beta )$ be the center of the circle $C_2$ of radius $5$ and $O_1(0,0)$ be the circle $C_1$ o0f radius $4$.

Then $O_{1}B=4$ and $O_{2}B=5$

$O_1{O}_2=\sqrt{25-16}=3=\sqrt{{\alpha }^2+{\beta }^2}$ ...(1)

Since slope of $O_{1}B$ is $\frac{3}{4}$ and $O_{1}B$ is $\perp$ to $O_1{O}_2$

$\therefore -\frac{\alpha }{\beta }=\frac{3}{4}\Rightarrow \alpha =\frac{-3\beta }{4}$ ...(2)

From (1) and (2), we get $9={\beta }^2+\frac{9{\beta }^2}{16}$

$\Rightarrow \frac{25}{16}{\beta }^2=9\Rightarrow {\beta }^2=\frac{144}{25}\Rightarrow \beta =\pm \frac{12}{5}$

when $\beta =\frac{12}{5};\alpha \Rightarrow \frac{-3}{4}\times \frac{12}{5}=\frac{-9}{5}$

and when $\beta =\frac{-12}{5};\alpha \Rightarrow \frac{-3}{4}\times \frac{-12}{5}=\frac{9}{5}$

Thus, the coordinate of the center of circle $C_2$ are

$(\frac{9}{5},\frac{-12}{5})$ or $(\frac{-9}{5},\frac{12}{5})$