JEE Main 26-Aug-2021 Shift 2 Solved Paper

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 ...(ii)
Since circles C and C1 intersect each other at PQ, so PQ is common chord.
∴ 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=√(

11

2

)2+(−6)2−5
=√

12

4

+36−5=√

245

4

=

√245

2

∴ Diameter of C=2× Radius =√245=7√5