Multiple choice

Tangents drawn from the point $P(1,8)$ to the circle $\displaystyle x^{2}+y^{2}-6x-4y-11=0$ touch the circle at the point A and B. The equation of the circumcentre of the $\displaystyle \triangle PAB$ is

  1. $\displaystyle x^{2}+y^{2}+4x-6y+19=0$
  2. $\displaystyle x^{2}+y^{2}-4x-10y+19=0$
  3. $\displaystyle x^{2}+y^{2}-2x+6y-29=0$
  4. $\displaystyle x^{2}+y^{2}+6x-4y+19=0$
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B Correct answer
Explanation

The circumcircle of triangle PAB, where P is an external point and A, B are points of tangency, has the segment OP as diameter, where O is the center of the circle. Center of circle x^2+y^2-6x-4y-11=0 is (3,2). P is (1,8). Diameter endpoints are (3,2) and (1,8). Midpoint (center) = (2,5). Radius squared = (3-2)^2 + (2-5)^2 = 1+9 = 10. Equation: (x-2)^2 + (y-5)^2 = 10 => x^2-4x+4 + y^2-10y+25 = 10 => x^2+y^2-4x-10y+19=0.

AI explanation

The centre of the given circle is (3, 2) and its radius is sqrt(24). The circumcircle of triangle PAB has PA and PB as chords; since PA is perpendicular to OA, the centre of the circumcircle is the midpoint of PO. The midpoint of P(1,8) and O(3,2) is (2, 5), which becomes the centre of the circumcircle, and its radius is the distance from this midpoint to P, which is sqrt(29). Thus, the equation of the circumcircle is (x - 2)^2 + (y - 5)^2 = 29, which simplifies to x^2 + y^2 - 4x - 10y + 19 = 0.

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