If the point of intersection of the tangents drawn at the points where the line $5 x+y+1=0$ cuts the circle…
If the point of intersection of the tangents drawn at the points where the line $5 x+y+1=0$ cuts the circle $x^2+y^2-2 x-6 y-8=0$ is $(\underline{a, b})$, then $5 a+b=$
3
-44
-1
4
Solution
Given circle,
$
x^2+y^2-2 x-6 y-8=0
$
So, chord of contact of two tangents drawn from the point $P(a, b)$ to the given circle is
$
\begin{gathered}
x a+y b-(x+a)-3(y+b)-8=0 \\
\Rightarrow \quad(a-1) x+(b-3) y-(a+3 b+8)=0
\end{gathered}
$
As, this line coincides with $5 x+y+1=0$
$
\begin{array}{ll}
\therefore \quad & \frac{a-1}{5}=\frac{b-3}{1}=\frac{-(a+3 b+8)}{1} \\
\Rightarrow \quad & a-1=-5(a+3 b+8) \text { and } \\
& b-3=-a-3 b-8 \\
\Rightarrow & 6 a+15 b=-39 \text { and } a+4 b=-5
\end{array}
$
On solving Eqs. (i) and (ii), we get
$
a=-9 \text { and } b=1
$
Hence, $5 a+b=-45+1=-44$
