JEE Main 17 March 2021 Shift 1 Solved Paper

Given, 2x−y+1=0 is a tangent to the circle at (2,5)
So, normal at (2,5) will be ‌

(y−5)

(x−2)

=‌

−1

2

⇒2y−10‌‌=−x+2
x+2y‌‌=12
Now, it is also given that centre lies on x−2y=4.
So, coordinates of centre will be the solution of
{

x+2y=12

x−2y=4

⇒x=8,y=2
Radius will be the distance between (8,2) and (2,5)
r2=(8−2)2+(2−5)2
⇒‌r2‌=36+9
⇒‌r‌=√45=3√5