Animated Solution for Mathematics - Circles: 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

To begin, we identify the properties of the given circle defined by the equation $x^2+y^2-6x-4y-11=0$. By comparing this to the general form $x^2+y^2+2gx+2fy+c=0$, we determine the coefficients $2g=-6$ and $2f=-4$.

This yields $g=-3$ and $f=-2$. Since the center $C$ is located at $(-g,-f)$, we find the center of the circle to be $C(3,2)$.

We consider the tangents drawn from point $P(1,8)$ to the circle, touching at points $A$ and $B$. A fundamental property of geometry dictates that the radius is perpendicular to the tangent at the point of contact. Thus, $CA\perp PA$ and $CB\perp PB$, implying $\angle PAC=90^{\circ }$ and $\angle PBC=90^{\circ }$.

Consider the quadrilateral $PACB$. The sum of the opposite angles $\angle PAC$ and $\angle PBC$ is $90^{\circ }+90^{\circ }=180^{\circ }$.

Because the opposite angles sum to $180^{\circ }$, the quadrilateral $PACB$ is cyclic. Consequently, all four points $P,A,C,$ and $B$ lie on the circumference of a single circle.

Since $\angle PAC=90^{\circ }$, the segment $PC$ must serve as the diameter of this circumcircle. This insight allows us to bypass the calculation of the coordinates of $A$ and $B$ entirely.

We now define the circle using the diameter form, where $P(1,8)$ and $C(3,2)$ are the endpoints of the diameter. The equation is given by:

Substituting the coordinates of $P$ and $C$ into the formula, we obtain:

Expanding the terms, we calculate the $x$ and $y$ components:

By grouping the terms and rearranging them into the standard form, we arrive at the final equation of the circumcircle: