Circle C touches x=2y at (2,1), then equation of circle C is given by (x−2)2+(y−1)2+λ(x−2y)=0 ⇒x2−4x+4+y2−2y+1+λx−2λy=0 ⇒x2+y2+(λ−4)x+(−2−2λ)y+5=0 Since circles C and C1 intersect each other at PQ, so PQ is common chor(d) ∴ Equation of PQ will be C−C1=0⇒x(λ−4)+y(−4−2λ)+10=0 Again PQ is diameter of circle C1. So, PQ passes through centre of C1. Now, equation of C1 is x2+y2+2y−5=0 ∴ Centre of C1 is (0,−1). So, PQ passes through (0,−1) ∴0(λ−4)−1(−4−2λ)+10=0⇒4+2λ+10=0⇒λ=−7 On putting λ=−7 in Eq. (i), we get equation of circle C as x2+y2−11x+12y+5=0 ∴ Radius of C=√(