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
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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 $
