How to find the two opposite end coordinates of the perpendicular diameter?

Question

If $x^2 + y^2 - x - 2y - 5 = 0$ is a circle equation and $(-1,-1),(2,3)$ are two opposite end of a diameter of this circle. Find the two opposite end coordinates of the perpendicular diameter.

Answer 1

Method 1:

The equation of the circle can be written as $$(x-\frac{1}{2})^2 + (y-1)^2 = (\frac{5}{2})^2$$ Hence, the center of the circle is $(\frac{1}{2}, 1)$

A line passing through $(-1, -1)$ and $(2, 3)$ has a slope of $\frac{4}{3}$

The slope of the perpendicular diameter will be the negative reciprocal of $\frac{4}{3}$, that is $-\frac{3}{4}$

A line passing through the center of the circle $(\frac{1}{2}, 1)$, with a slope of $-\frac{3}{4}$ has an equation $$y = -\frac{3}{4}x + \frac{11}{8}$$ Substitute this into the equation of the circle to get $$x^2 - x - \frac{15}{4} = 0$$ $$\implies x \in \left\{\frac{5}{2}, -\frac{3}{2}\right\}$$ $$\implies y \in \left\{-\frac{1}{2}, \frac{5}{2}\right\}$$ The required coordinates are $(\frac{5}{2},-\frac{1}{2})$ and $(-\frac{3}{2},\frac{5}{2})$

Method 2:

The center of the circle is the midpoint of $(-1, -1)$ and $(2, 3)$, that is $(\frac{1}{2}, 1)$

The length of the circle's diameter is the distance between $(-1, -1)$ and $(2, 3)$, that is $5$

The slope of the perpendicular diameter will be the negative reciprocal of $\frac{4}{3}$, that is $-\frac{3}{4}$

The two endpoints of the diameter must be of the form $(\frac{1}{2}+t, 1-\frac{3}{4}t)$ and $(\frac{1}{2}-t, 1+\frac{3}{4}t)$

The distance between the endpoints should be equal to the diameter of the circle $$\sqrt{(2t)^2 + (\frac{3}{2}t)^2} = 5$$ $$\implies t = 2$$

The required coordinates are $(\frac{5}{2},-\frac{1}{2})$ and $(-\frac{3}{2},\frac{5}{2})$