Tangents drawn from the point P(1, 8) to the circle touch the circle at the points A and B. The equation of the circumcircle of the triangle is [2009]

Question

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 $P A B$ is [2009]

Smart Achievers · Accepted Answer

(2)

Given that tangents PA and PB are drawn from the point P(1, 3) to circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ with centre C(3, 2)

Clearly the circumcircle of $∆ P A B$ will pass through C and as $∠ A = 90 °$ , PC must be a diameter of the circle.

$∴$ Equation of required circle is

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

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

Solution

Q.
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 $P A B$ is [2009]

1 $x^{2} + y^{2} + 4 x - 6 y + 19 = 0$

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

3 $x^{2} + y^{2} - 2 x + 6 y - 29 = 0$

4 $x^{2} + y^{2} - 6 x - 4 y + 19 = 0$

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Solution

Q. Tangents drawn from the point P(1, 8) to the circle x2+y2-6x-4y-11=0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is [2009]

1 x2+y2+4x-6y+19=0

2 x2+y2-4x-10y+19=0

3 x2+y2-2x+6y-29=0