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Q.
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.
Conic Sections
Solution:
Given equation is $x^{2}+y^{2}-3x+5y-4=0 \, \ldots\left(i\right)$
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 $\left(1, -6\right)$.
Since $C$ is mid-point of $AB$,