Question

A tangent to the circle $x^2+y^2=1$ through the point $(0,5)$ cuts the circle $x^2+y^2=4$ at $A$ and $B$. The tangent to the circle $x^2+y^2=4$ at $A$ and $B$ meets at $C$. The coordinates of $C$ are

• A. $\left(\frac{8}{5}\sqrt{6},\frac{4}{5}\right)$
• B. $\left(\frac{8}{5}\sqrt{6},-\frac{4}{5}\right)$
• C. $(-\frac{8}{5}\sqrt{6},-\frac{4}{5})$
• D. none of these

Answer 1

The correct option is A $\left(\frac{8}{5}\sqrt{6},\frac{4}{5}\right)$

If $C\equiv (\alpha ,\beta )$ then chord of contact of the tangents from it to the circle $x^2+y^2=4$ is $\alpha x+\beta y=4$ which passes through the point $(0,5)$
$\therefore 5\beta =4\Rightarrow \beta =\frac{4}{5}$
Also, $ax+\beta y=4$ is tangent to the circle $x^2+y^2=1$
$\therefore \frac{4}{\sqrt{{\alpha }^2+{\beta }^2}}=1\Rightarrow 16={\alpha }^2+{\left(\frac{4}{5}\right)}^2$
$\Rightarrow {\alpha }^2=16-\frac{16}{25}=\frac{384}{25}\therefore \alpha =\pm \frac{8\sqrt{6}}{5}$
Therefore point C is $(\frac{8\sqrt{6}}{5},\frac{4}{5}),(\frac{-8\sqrt{6}}{5},\frac{4}{5})$