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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 $5 a+b=$
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Verified Answer
The correct answer is:
-44
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, $5 a+b=-45+1=-44$
$$
x^2+y^2-2 x-6 y-8=0
$$
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, $5 a+b=-45+1=-44$
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