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# 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.

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