JEE Advanced Model Paper 18 with solutions for online practice

Let P (x1,y1) & Q (x2,y2) then the roots of the equation (mx+1)2 - 4ax = 0
x2+(

3m−4a

m2

) x +

1

m2

= 0 are x1 and x2
And the roots of the equation y2 = 4a (

y−1

m

) = y2−

4a

m

y +

4a

m

= 0 and y1 and y2
Required circle is
x2+y2 + (

2m−4a

m2

) X + (−

4a

m

) y +

1

m2

+

4a

m

= 0 ..... (1)
x2+y2 + 18x + 4y + 5 = 0 ..... (2)
From (1) and (2)

−4a

m

= 4 ,

1

m2

+

4a

m

= 5 and

3m−4a

m2

= 10
⇒ m =

1

3

and a = −

1

3