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  4. If the coordinates of one end of a diameter of the circle x2+y2+4x-8y+5=0, is (2,1), the coordinates of the other end is

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Q. If the coordinates of one end of a diameter of the circle $x^2+y^2+4x-8y+5=0$, is (2,1), the coordinates of the other end is

WBJEEWBJEE 2011

Solution:

Since, one end of a diameter of the circle is $(2,1)$.
Let the other end of a circle is $(h, k)$.
But centre of a circle is $(-2,4)$.
$\therefore \frac{h+2}{2}=-2$ and $\frac{k+1}{2}=4$
$\Rightarrow h=-6$ and $k=7$
Hence, required point is $(-6,7)$.