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
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Answer: 3
Solution:
Formula: Equation of Tangent in Point form
Equation of circle with PR as diameter is $(x+3)(x-\alpha)+(y-2)(y-4)=0$
$\mathrm{Q}(9,10)$ 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 $\mathrm{R}=8 x+y-108=0$
From (1), (2) $(x, y)=(\frac{25}{2}, 8)$
$\Rightarrow 2 \mathrm{x}-\mathrm{ky}=1 \Rightarrow 25-8 \mathrm{k}=1 \Rightarrow 8 \mathrm{k}=24 \Rightarrow \mathrm{k}=3$
