We express the given circle \(2 x^{2}+2\left(y-\frac{3}{2}\right)^{2}=2\) in the form \((x-a)^{2}+(y-b)^{2}=r^{2} \ldots \ldots .(1)\) \((x-a)^{2}+(y-b)^{2}=r^{2} \ldots \ldots \ldots .\) \((a . b)\) as centre and radius \(r\)
\(2 x^{2}+2\left(x-\frac{3}{2}\right)^{2}=2\)
We divide through by 2
\(x^{2}+\left(y-\frac{3}{2}\right)^{2}=1\)
this is the same as
\((x+0)^{2}+\left(y-\frac{3}{2}\right)^{2}=12\)omparing (1) and (2)
\(a=0 . b=-\frac{3}{2} \cdot r=1\)
\(\Rightarrow\left(0,-\frac{3}{2}\right)\) and 1