If the coordinates of point, of contact of the circles $\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+8…
If the coordinates of point, of contact of the circles $\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+8 \mathrm{y}+4=0$ and $\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}=0$ is $(\mathrm{a}, \mathrm{b})$, then $a+2 b=$
-1
-2
0
1
Solution
Circle $x^2+y^2-4 x+8 y+4=0$
have centre $P(2,-4)$ and radius 4 unit and circle $x^2+y^2$ $+2 x=0$ have centre $Q(-1,0)$ and radius 1 unit
Let $O(a, b)$ be the point of contact.
$
\begin{aligned}
& O(a, b)=\frac{4(-1)+1(2)}{4+1}, \frac{4(0)+1(-4)}{4+1} \\
& =\left(-\frac{2}{5},-\frac{4}{5}\right) \\
& \therefore a+2 b=-2
\end{aligned}
$
