Points P(-3,2),Q(9,10) and R(α ,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…

Let $S\equiv ({x}_{1},{y}_{1})$S≡(x1​,y1​)

Given:

$$P\equiv (-3,2) Q\equiv (9,10) R\equiv (\alpha ,4)$$P≡(−3,2)Q≡(9,10)R≡(α,4)

Since, $PQ\perp RQ$PQ⊥RQ, so

$${m}_{PQ}\times {m}_{QR}=-1$$mPQ​×mQR​=−1

$$\Rightarrow (\frac{10-2}{9+3})\times (\frac{10-4}{9-\alpha })=-1$$⇒(9+310−2​)×(9−α10−4​)=−1

$$\Rightarrow (\frac{1}{9-\alpha })=-\frac{1}{4}$$⇒(9−α1​)=−41​

$$\Rightarrow \alpha =13$$⇒α=13

So, $R\equiv (13,4)$R≡(13,4)

So, centre of circle is $O\equiv (\frac{13-3}{2},\frac{4+2}{2})\equiv (5,3)$O≡(213−3​,24+2​)≡(5,3)

Also,

$${m}_{OQ}\cdot {m}_{QS}=-1$$mOQ​⋅mQS​=−1

$$\Rightarrow (\frac{3-10}{5-9}){m}_{QS}=-1$$⇒(5−93−10​)mQS​=−1

$$\Rightarrow {m}_{QS}=-\frac{4}{7}$$⇒mQS​=−74​

Equation of $QS$QS is

$$y-10=-\frac{4}{7}(x-9)$$y−10=−74​(x−9)

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

Also,

$${m}_{OR}\cdot {m}_{RS}=-1$$mOR​⋅mRS​=−1

$$\Rightarrow (\frac{4-3}{13-5}){m}_{RS}=-1$$⇒(13−54−3​)mRS​=−1

$$\Rightarrow {m}_{RS}=-8$$⇒mRS​=−8

Equation of $RS$RS

$$y-4=-8(x-13)$$y−4=−8(x−13)

$$\Rightarrow 8x+y=108\ldots (2)$$⇒8x+y=108…(2)

Solving equations (1)&(2), we get

$${x}_{1}=\frac{25}{2};{y}_{1}=8$$x1​=225​;y1​=8

Since, $S({x}_{1},{y}_{1})$S(x1​,y1​) lies on $2x-ky=1$2x−ky=1

$$25-8k=1$$25−8k=1

$$\Rightarrow 8k=24$$⇒8k=24

$$\Rightarrow k=3$$⇒k=3