### Circle Question 5

#### Question 5 - 29 January - Shift 2

A circle with centre $(2,3)$ and radius 4 intersects the line $x+y=3$ at the points $P$ and $Q$. If the tangents at $P$ and $Q$ intersect at the point $S(\alpha, \beta)$, then $4 \alpha-7 \beta$ is equal to

## Show Answer

#### Answer: 11

#### Solution:

#### Formula: Chord of contact TT’ of two tangents

The given circle is $ x^2+y^2-4 x-6 y-3=0 $

Chord of contact

$ \alpha x+\beta y-2(x+\alpha)-3(y+\beta)-3=0 $

$\Rightarrow(\alpha-2) x+(\beta-3) y-(2 \alpha+3 \beta+3)=0 $

$\because$ But the equation of chord of contact is given as $ x+y-3=0$

comparing the coefficients $ \frac{\alpha-2}{1}=\frac{\beta-3}{1}=-\left(\frac{2 \alpha+3 \beta+3}{-3}\right) $

On solving $\alpha=-6$, $ \beta=-5 $

Now $ 4 \alpha-7 \beta=11 $