Question

The centre of a circle is $C(4,5)$ and one end of a diameter is $A(3,2)$, 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 $A(3,2)$

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

Since the centre $C(4,5)$ of the circle is the mid-point of $AB$,

$\frac{\alpha +3}{2}=4$ and $\frac{\beta +2}{2}=5$

$\Rightarrow \alpha +3=8$ and $\beta +2=10$

$\Rightarrow \alpha =8-3=5$ and $\beta =10+2=8$

$\therefore$The coordinates of the other end of the diameter are $(5,8)$