How do I find the equation of the circumcircle?

Question

Tangents are drawn from the point P(1,8) to the circle $x^2+y^2-6x-4y-11=0$ touch the circle at points A and B.So what will be the equation of the circumcircle of $\Delta$ PAB?I am not asking you to to solve the question and tell me the answer.Just tell me how to do it please.

Answer 1

Hint:

Equation of chord is given by $T = 0$:

$$x +8y-3(x+1)-2(y+8)-11=0$$

Solving it with circle equation gives points $A$ and $B$.

Now consider the equation of required circle be:

$$x^2++y^2+2gx+2fy+c=0$$

This equation satisfies $A$, $B$, $P$, so you can find $g,f$ and $c$.

Answer 2

General approach:

Step 1: Find the coordinates of the points $A$ and $B$.

Step 2: Find the equation of the circle through $P$, $A$, $B$.

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For step 1, letting $M$ be the center of the circle, note that $PAMB$ is a square. (Why?) Since you know the coordinates of $P$ and $M$, you can figure out the coordinates of $A$ and $B$.

For step 2, use the general equation of a circle $x^2 + 2Px + y^2 + 2Qy + C = 0$ and plug in the coordinates for $P$, $A$ and $B$. You will get a system of linear equations and will be able to solve for $P$, $Q$ and $C$.