Points P (-3,2), Q (9,10) and R (α, 4) lie on a circle C with P R as its diameter. tangents to C at

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10-1.Circle and System of Circles

hard

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

$3$

$6$

$9$

$12$

(JEE MAIN-2023)

Solution

$m _{ PQ } \cdot m _{ QR }=-1$

$\Rightarrow \frac{10-2}{9+3} \times \frac{10-4}{9-\alpha}=-1 \Rightarrow \alpha=13$

$m _{ op } \cdot m _{ Qs }=-1 \Rightarrow m _{ Qs }=-\frac{4}{7}$

Equation of $QS$

$y-10=-\frac{4}{7}(x-9)$

$\Rightarrow 4 x+7 y=106 \ldots(1)$

$m_{O R} \cdot m_{R S}=-1 \Rightarrow m_{R S}=-8$

Equation of RS

$y-4=-8(x-13)$

$\Rightarrow 8 x+y=108$

Solving eq. $(1)$ and $(2)$

$x _1=\frac{25}{2} y _1=8$

$S \left( x _1, y _1\right)$ lies on $2 x – ky =1$

$25-8 k =1$

$\Rightarrow 8 k =24$

$\Rightarrow k =3$

Standard 11

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