The centre of a circle is $( -6,\ 4)$. If one end of the diameter of the circle is at $( -12,\ 8)$, then find the point of the other end.

Given: The centre of a circle is $( -6,\ 4)$. If one end of the diameter of the circle is at $( -12,\ 8)$.

To do: To find the point of the other end.

Solution:

As given, the center of the circle $O=( -6,\ 4)$

One end of the diameter of the circle $A=( -12,\ 8)$

Let $B( x,\ y)$ be the another end of the diameter of the circle.

Then $OA=OB$,

$\Rightarrow ( -6,\ 4)=( \frac{-12+x}{2},\ \frac{8+y}{2})$          [On using mid point formula]

$\Rightarrow \frac{x-12}{2}=-6$ and $\frac{y+8}{2}=4$

$\Rightarrow x-12=-12$ and $y+8=8$

$\Rightarrow x=0$ and $y=0$

Thus, $B( 0,\ 0)$ is the another point of the diameter of the circle.