The line $x + 3y = 20$ intersects the circle $x^2 + y^2 - 6x - 8y = 0$ at the points $P$ and $Q$ - Leaving Cert Mathematics - Question 5 - 2011

The midpoint $C$ of the points $P(-1, 7)$ and $Q(-1, 4)$ can be calculated as:

$C \left( \frac{-1 + (-1)}{2}, \frac{7 + 4}{2} \right) = C\left(-1, \frac{11}{2}\right)$

The radius $r$ of the circle whose diameter is $[PQ]$ is half of the distance $d$ between the points $P$ and $Q$:

\Rightarrow r = \frac{d}{2} = \frac{3}{2}$$

The general equation of a circle is given by:

$(x - h)^2 + (y - k)^2 = r^2$ where $(h, k)$ is the center and $r$ is the radius. Thus, the equation for our circle becomes:

$\left(x + 1\right)^2 + \left(y - \frac{11}{2}\right)^2 = \left(\frac{3}{2}\right)^2$

Substituting the values:

$\left(x + 1\right)^2 + \left(y - \frac{11}{2}\right)^2 = \frac{9}{4}$