CircleEasy
Question
The tangent and normal at the point P (18, 12) of the parabola y2 = 8x intersects the x-axis at the points A and B respectively. The equation of the circle through P, A and B is given by:-
Options
A.x2 + y2 + 4x - 540 = 0
B.x2 + y2 - 6x - 360 = 0
C.x2 + y2 - 4x - 396 = 0
D.x2 + y2 - 2x - 444 = 0
Solution
We know that
PS = AS = SB
⇒ S is the circum-centre of ᐃPAB
∴ Equation of the required circle is
(x - 2)2 + (y - 0)2 = (2 - 18)2 + (0 - 12)2
⇒ x2 + y2 - 4x - 396 = 0
PS = AS = SB
⇒ S is the circum-centre of ᐃPAB
∴ Equation of the required circle is
(x - 2)2 + (y - 0)2 = (2 - 18)2 + (0 - 12)2
⇒ x2 + y2 - 4x - 396 = 0
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