The correct option is D p2>8q2
Let B(h, 0) be the midpoint of the chord drawn from point A(p, q).
Also, the center of the circle is C(p2,q2).
Then, we have BC⊥AB. Therefore,
(q2)−0(p2)−h(q−0p−h)=−1 ...(Product of slopes of two perpendicular lines, m1.m2=−1)∴(qp−2h)(q−0p−h)=−1∴2h2−3ph+p2+q2=0
Since two such chords exist, the above equation must have two distinct real roots, i.e.,
Discriminant > 0