The ends of a diameter of a circle have the coordinates (3, 0) and (-5, 6). PQ is another diameter where Q has the coordinates (-1, -2). Find the coordinates of P and the radius of the circle.

Let AB be the diameter where coordinates of A are (3, 0) and of B are (-5, 6).

∴ Coordinates of its midpoint will be $\Big(\dfrac{3 + (-5)}{2}, \dfrac{0 + 6}{2}\Big)$ or (-1, 3).

Now PQ is another diameter in which the coordinates of Q are (-1, -2).

Let coordinates of P be (x, y), then by mid-point formula coordinates of mid-point will be $\Big(\dfrac{-1 + x}{2}, \dfrac{-2 + y}{2}\Big)$

Since, the diameters of circle intersect at their midpoint.

$\therefore \dfrac{-1 + x}{2} = -1 \text{ and } \dfrac{-2 + y}{2} = 3 \\[1em] \Rightarrow -1 + x = -2 \text{ and } -2 + y = 6 \\[1em] \Rightarrow x = -2 + 1 \text{ and } y = 6 + 2 \\[1em] \Rightarrow x = -1 \text{ and } y = 8.$

∴ Coordinates of P will be (-1, 8).

Radius = OP, by distance-formula we get,

$OP = \sqrt{(-1 - (-1))^2 + (8 - 3)^2} \\[1em] = \sqrt{(-1 + 1)^2 + (5)^2} \\[1em] = \sqrt{0^2 + 25} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Hence, the radius of circle is 5 units and the coordinates of P is (-1, 8).