Question

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

• A. $(2,-6)$
• B. $(1,-6)$
• C. $(2,6)$
• D. $(1,6)$

Answer 1

Answer: (B)

Given equation is $x^2+y^2-3x+5y-4=0\dots \left(i\right)$
It can be easily seen that $\left(i\right)$
represent a circle with centre
$C\left(\frac{3}{2},-\frac{5}{2}\right)\cdot$
Let $A\left(2,1\right)$ and $B\left(\alpha ,\beta \right)$ be the ends of the diameter.
Since $C$ is mid-point of $AB$,
$\therefore \frac{\alpha +2}{2}=\frac{3}{2}$ and $\frac{\beta +1}{2}=-\frac{5}{2}$
$\Rightarrow \alpha =1$ and $\beta =-6$.
Hence, the other end of the diameter is $\left(1,-6\right)$.