JEE Main 17 March 2021 Shift 2 Solved Paper

Given, equation of circle is
x2+y2=25..(i)
∴ Equation of tangent to the given circle at R(3,4) is given by
3x+4y=25...(ii)
(by the rule, xx1+yy1=25⇒3x+4y=25 )
For OP, we must put y=0 and for OQ, put x=0 in Eq. (ii). ∴‌‌OP=‌

25

3

and OQ=‌

25

4

∴‌‌PQ=‌

125

12

∵OP+PQ+QO=25
∴ In centre of According to the question, incentre of △OPQ= centre of the required circle
r= radius of the required circle which is passing through origin.
=√(‌

25

12

−0)2+(‌

25

12

−0)2
⇒ r2=2(‌

25

12

)2
⇒ r2=‌

625

72