# Write equation of circle whose diametrically opposite point are (1,2) & (-1,-2) is?

$\begin{array}{1 1}(a)\;x^2-y^2=5\\(b)\;x^2+y^2=5\\(c)\;x+y=5\\(d)\;2x^2-3y^2=5\end{array}$

Given diametrically opposite point $(1,2),(-1,-2)$
Hence centre of the circle will be mid-point of the diametrical oposite point.
Mid point of $(1,2)$ and $(-1,2)$, $(h,k)$ is given by
$h=\large\frac{1+(-1)}{2}$$=0 k=\large\frac{2+(-2)}{2}$$=0$
$\therefore$ The centre of the circle is $=(0,0)$ and
diameter is $\sqrt {1+1)^2+(2+2)^2}=\sqrt {20}$
$\Rightarrow\:(radius)^2=r^2=(\large\frac{\sqrt {20}}{2})^2$$=5$
$\therefore\:$ Equation of the circle is
$(x-0)^2+(y-0)^2=5$
$x^2+y^2=5$
Hence (b) is the correct answer.
edited Mar 24, 2014