Question

The centre of a circle is $C(2,6)$ and one end of a diameter is $A(3,5)$, find the coordinates of the other end.

Answer 1

Let the other end of the diameter of the given circle be $B(\alpha ,\beta )$ whose one end is $A(3,5)$

Midpoint of $AB=(\frac{\alpha +3}{2},\frac{\beta +5}{2})$

Since the centre $C(2,6)$ of the circle is the mid-point of $AB$,

$\frac{\alpha +3}{2}=2$ and $\frac{\beta +5}{2}=6$

$\Rightarrow \alpha +3=4$ and $\beta +5=12$

$\Rightarrow \alpha =4-3=1$ and $\beta =12-5=7$

$\therefore$The coordinates of the other end of the diameter are $(1,7)$