### Circle Question 3

#### Question 3 - 25 January - Shift 2

Points $P(-3,2), Q(9,10)$ and $R(\alpha, 4)$ lie on a circle $C$ with PR as its diameter. The tangents to $C$ at the points $Q$ and $R$ intersect at the point $S$. If $S$ lies on the line $2 x-k y=1$, then $k$ is equal to

## Show Answer

#### Answer: 3

#### Solution:

#### Formula: Equation of Tangent in Point form

Equation of circle with PR as diameter is $(x+3)(x-\alpha)+(y-2)(y-4)=0$

$Q(9,10)$ lies on the circle $\Rightarrow \alpha=13$

equation of circle $x^2+y^2-10 x-6 y-31=0$

Equation of tangent at $Q: 4 x+7 y-106=0$

Equation of tangent $R=8 x+y-108=0$

From (1), (2) $(x, y)=(\frac{25}{2}, 8)$

$\Rightarrow 2 x-ky=1 \Rightarrow 25-8 k=1 \Rightarrow 8 k=24 \Rightarrow k=3$