‌x2+y2=3‌ and ‌x2=2y ‌y2+2y−3=0⇒(y+3)(y−1)=0 ‌y=−3‌ or ‌y=1 ‌y=1x=√2⇒P(√2,1) 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,α) coordinates R1=2√3‌ (Given ‌ Line L act as tangent Apply P = r (condition of tangency) ‌⇒|‌
α−3
√3
|=2√3 ‌⇒|α−3|=6 α−3=6‌‌‌ or ‌‌‌α−3=−6 ⇒α=9‌‌α=−3 ‌△PQ1Q2=‌