Circle
Tangents and Circumcircle
JEE Advanced 2009 Paper 1
Grade 11
Question:
Tangents drawn from the point $P(1, 8)$ to the circle $x^2 + y^2 - 6x - 4y - 11 = 0$ touch the circle at the points $A$ and $B$. The equation of the circumcircle of triangle $PAB$ is:
(1) $x^2+y^2-4x-10y+19 = 0$
(2) $x^2+y^2+4x-10y+19 = 0$
(3) $x^2+y^2-2x+6y-29 = 0$
(4) $x^2 + y^2 - 6x + 4y + 19 = 0$
Step-by-Step Solution
Key Concept: Write the chord of contact from $P(1, 8)$: $T = x + 8y - 3(x + 1) - 2(y + 8) - 11 = 0$, giving $x - 3y + 15 = 0$. Use the family $S + \lambda T = 0$, substitute $P$ to find $\lambda = 2$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)