Given: x2+y2=3 and x2=2y
⇒y2+2y−3=0
⇒(y+3)(y−1)=0
⇒y=1,−3(Rejected as it gives imaginary value ofx)
⇒y=1
⇒x=2
⇒P≡(2,1)
Now, P lies on the line 2x+y=α.
⇒2(2)+1=α
⇒α=3
For circle C1, Q1 lies on y−axis.
Let Q1≡(0,β) and R1=23 (given)
Line L act as tangent so applying the condition of tangency
⇒P=r
⇒∣3β−3∣=23
⇒∣β−3∣=6
⇒β−3=6,−6
⇒β=9,−3
So, {Q}_{1}\equiv (0,9)&{Q}_{2}\equiv (0,-3)
⇒Ar(ΔPQ1Q2)=21∣20019−3111∣
⇒Ar(ΔPQ1Q2)=21(2(12))
⇒Ar(ΔPQ1Q2)=62
⇒(ΔPQ1Q2)2=72