If the line $4 x+4 y-11=0$ intersects the circle $x^2+y^2-4 x-6 y+4=0$ at $A$ and $B$, then the point of…
If the line $4 x+4 y-11=0$ intersects the circle $x^2+y^2-4 x-6 y+4=0$ at $A$ and $B$, then the point of intersection of the tangents drawn at $A, B$ is
$(-1,2)$
$(-1,-2)$
$(2,1)$
$(-2,-1)$
Solution
Let required point of intersection is $\left(x_1, y_1\right)$, the equation of chord of contact with respect to given circle $x^2+y^2-4 x-6 y+4=0$ is
$
\begin{aligned}
& x x_1+y y_1-2\left(x+x_1\right)-3\left(y+y_1\right)+4=0 \\
& \Rightarrow\left(x_1-2\right) x+\left(y_1-3\right) y+\left(4-3 y_1-2 x_1\right)=0
\end{aligned}
$
Let the Eq. (i) represent the line $A B, 4 x+4 y-11=0$ itself, then
$
\frac{x_1-2}{4}=\frac{y_1-3}{4}=\frac{4-3 y_1-2 x_1}{-11}=K \text { (let) }
$
Then,
$
\begin{aligned}
& x_1=4 K+2 \\
& y_1=4 K+3
\end{aligned}
$
And $2 x_1+3 y_1-4=11 K$
From Eq. (ii)
$
\begin{array}{rlrl}
8 K+4+12 K+9-4 & =11 K \\
& & 9 K+9 & =0 \\
\Rightarrow & & K & =-1, \\
& \text { So, } & \left(x_1, y_1\right) & =(-2,-1) .
\end{array}
$