IIT-JEE 2009 · Paper 1 · Mathematics Q22

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MathematicsSingle correct+3 / -1Question 22 of 60

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 the triangle $PAB$ is

x^2+y^2+4x-6y+19=0

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

x^2+y^2-2x+6y-29=0

x^2+y^2-6x-4y+19=0

Answer and worked solution

Official answer: B

Right angles. The circle has centre $C(3,2)$ . The radii CA and CB are perpendicular to the tangents PA and PB, so A and B lie on the circle with diameter PC; that circle is the circumcircle of PAB.

Diameter form with $P(1,8)$ , $C(3,2)$ :
$(x-1)(x-3)+(y-8)(y-2)=0\implies x^2+y^2-4x-10y+19=0.$

(B)

Topics:Tangents and normals Equation of a circle

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IIT-JEE 2009 · Paper 1, Mathematics question 22