Explanation:
A Given,
Equation tangent to parabola at \((1,2)\) is,
\(y-2=x-1\)
\(x-y+1=0\)
Therefore equation of normal through \((1,2)\) will be,
\(\mathrm{x}+\mathrm{y}-3=0\)
Since, the centre lies on the normal
Let the coordinates of the centre be \((3-r, r)\)
Now, distance between \((1,2)\) and \((3-r, r)\) is \(r\)
\(\therefore \quad (3-r-1)^2+(r-2)^2=r^2\)
\(\sqrt{2}(r-2)= \pm r\)
\(r=\frac{2 \sqrt{2}}{\sqrt{2} \pm 1}\)
\(r=2 \sqrt{2}(\sqrt{2}-1)\)
\(r=4-2 \sqrt{2}(\text { since radius }\lt 3)\)
So, area of circle \(=\pi \mathrm{r}^2\)
\(=\pi(4-2 \sqrt{2})^2\)
\(=\pi(16+8-16 \sqrt{2})\)
\(=8 \pi(3-2 \sqrt{2})\)