JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Circle

Tangents drawn from the point

P (1, 8)

to the circle

x^{2} + y^{2} - 6 x - 4 y - 11 = 0

touch the circle at the points

A

and

B

. The equation of the circumcircle of the triangle

P A B

(A) $x^{2} + y^{2} + 4 x - 6 y + 19 = 0$
(B) $x^{2} + y^{2} - 4 x - 10 y + 19 = 0$ ✓ Correct
(C) $x^{2} + y^{2} - 2 x + 6 y - 29 = 0$
(D) $x^{2} + y^{2} - 6 x - 4 y + 19 = 0$

Answer: (B)

Solution

The centre of the given circle is

C (3, 2)

. Since

C A

and

C B

are perpendicular to

P A

and

P B

respectively (radius perpendicular to tangent),

C P

is the diameter of the circumcircle of triangle

P A B

. The equation of the circle with diameter

C P

where

C (3, 2)

and

P (1, 8)

is:

(x - 3) (x - 1) + (y - 2) (y - 8) = 0

x^{2} + y^{2} - 4 x - 10 y + 19 = 0

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option

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