Question:
If $(-1,3)$ and $(\alpha, \beta)$ are the extremities of the diameter of the circle $x^{2}+y^{2}-$ $6 x+5 y-7=0$, find the coordinates $(\alpha, \beta)$
Solution:
Given $x^{2}+y^{2}-6 x+5 y-7=0$
Centre $\left(3,-\frac{5}{2}\right)$
As $(-1,3) \&(\alpha, \beta)$ are the 2 extremities of the diameter, using mid - point formula we can write
$\frac{\propto-1}{2}=3$
$\Rightarrow \propto=7$
and $\frac{\beta+3}{2}=-\frac{5}{2}$
$\Rightarrow \beta=-8$
$(\alpha, \beta)=(7,-8)$
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