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One end of a diameter of the circle x^2+...

One end of a diameter of the circle `x^2+y^2-3x+5y-4=0` is (2,1). Find the coordinates of the other end.

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Centre of the circle is `(3/2, - 5/2)`. Let B`(alpha,beta)` be the other end of the diameter.
`therefore (alpha+2)/2=3/2` and `(beta+1)/2 = - 5/2 implies alpha=1,beta=-6`
`therefore` other end is (1,-6).
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