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

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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.

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