Equation of normal at (1, 2) on y² = 4x will be y – y¹ = $\frac{-{\mathrm{y}}_1}{2\mathrm{a}}(\mathrm{x}–{\mathrm{x}}_1)$

 $\Rightarrow \text{y }–\text{ 2 }=\frac{-2}{2·1}(\text{x }–\text{ 1})$

⇒ y – 2 = – x + 1 ⇒ x + y = 3                            

$\because$ It is tangent to (x – 3)² + (y + 2)² = r².

$\therefore$ Perpendicular distance from centre (3, – 2) = radius

$\Rightarrow \frac{|3-2-3|}{\sqrt{2}}\mathrm{ }=\text{ r}\Rightarrow \text{r }=\mathrm{ }\sqrt{2}\mathrm{ }\Rightarrow {\text{r}}^{\text{2}}\mathrm{ }=\text{ 2}.$