QJEEAdv20Mad374: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Mad374 Solution: Let O be the centre of the circle x 2 + y 2 = r 2 , where r > 5 …

As we know that the equation of family of circles passes through the points of intersection of given circle x² + y² = r² and line PQ : 2x + 4y = 5 is,

(x² + y² $$-$$ r²) + $$\lambda$$(2x + 4y $$-$$ 5) = 0 ......(i)

Since, the circle (i) passes through the centre of circle

x² + y² = r²,

So, $$-$$ r² $$-$$ 5$$\lambda$$ = 0

or 5$$\lambda$$ + r² = 0 ....(ii)

and the centre of circle (i) lies on the line x + 2y = 4, so centre ($$-$$ $$\lambda$$, $$-$$ 2$$\lambda$$) satisfy the line x + 2y = 4.

Therefore, $$-$$$$\lambda$$ $$-$$4$$\lambda$$ = 4

$$\Rightarrow$$ $$-$$5$$\lambda$$ = 4

$$\Rightarrow$$ r² = 4 {from Eq. (ii)}

$$\Rightarrow$$ r = 2