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  4. The other end of the diameter through the point (-1,1) on the circle x2+y2-6 x+4 y-12=0 is

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Q. The other end of the diameter through the point $ (-1,1) $ on the circle $x^{2}+y^{2}-6 x+4 y-12=0$ is

Bihar CECEBihar CECE 2012

Solution:

Given, circle is $x^{2}+y^{2}-6 x+4 y-12=0$
Centre of this circle is $(3,-2)$
Let other end of the diameter is $(\alpha, \beta)$.
$\therefore \frac{\alpha-1}{2}=3, \frac{\beta+1}{2}=-2$
$\Rightarrow \alpha=7,\, \beta=-5$
$\therefore $ Other end of the diameter is $(7,-5)$.