The centre of a circle is (–6,4). A diameter of the circle has its one end at the origin. Find its other end.

Let the diameter be OA where O is the origin (0, 0). Let the other end A be \$(x_1, y_1)\$.

The mid point of the diameter is the centre of the circle. So the mid point of OA is the centre \$(–6,4)\$.

But by the mid point formula, the mid point of OA is

\begin{align} \require{cancel} && \Bigl( \frac{ 0 + x_1 }{2}, \frac{ 0 + y_1 }{2} \Bigr) &= \Bigl( \frac{x_1}{2}, \frac{y_1}{2} \Bigr) \\ \end{align}

But this point is the centre (–6,4). So

\begin{align} && \Rightarrow \frac{x_1}{2} &= -6, & \frac{y_1}{2} &= 4 \\ && \Rightarrow x_1 &= -12, & y_1 &= 8 \\ \end{align}

Answer

The other end of the diameter is \$(x_1,y_1)=(−12,8)\$.