The correct option is A x2+y2−20x+8y−12=0
Normal at (at21,2at1) meets the parabola again at (at22,2at2)
Then, t2=−t1−2t1
Given parabola is y2=8x
⇒a=2
We are given P=(2,4)
⇒2at1=4
∴t1=1
⇒t2=−t1−2t1
⇒t2=−3
Then, P=(2,4) and Q=(18,−12)
Then, equation of circle is given by (x−2)(x−18)+(y−4)(y+12)=0
∴x2+y2−20x+8y−12=0