Question

If the length of the tangent drawn from $(\alpha ,\beta )$ to the circle $x^2+y^2=6$ be the length of the tangent from the same point to the circle $x^2+y^2+3x+3y=0$, then prove that ${\alpha }^2+{\beta }^2+4\alpha +4\beta +2=0$.

Answer 1

Misprint actually we have to prove ${\alpha }^2+{\beta }^2+2\alpha +2\beta +2\alpha \beta =0$

Given:

$\sqrt{s_1}=\sqrt{s_1}$

($\because$ length of tangent $=\sqrt{s_1}$)

$\Rightarrow \sqrt{{\alpha }^2+{\beta }^2-6}=\sqrt{{\alpha }^2+{\beta }^2+3\alpha +3\beta }$

$\Rightarrow {\alpha }^2+{\beta }^2-6={\alpha }^2+{\beta }^2+3\alpha +3\beta$

$\Rightarrow 3(\alpha +\beta )=-6$

$\Rightarrow \alpha +\beta =-2\to \left(1\right)$

To find

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha +2\beta +2\alpha \beta$

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta +2(\alpha +\beta )$

$\Rightarrow (\alpha +\beta {)}^2+2(\alpha +\beta )$

Using $\left(1\right)$

$\Rightarrow (-2{)}^2+2(-2)$

$\Rightarrow 4-4=0$

Hence proved.