The image of the point (3, 5) in the line x - y + 1 = 0 , lies on :

The slope of $PP ' = - 1$ ( As, $PP '$ is perpendicular to the given line)

$⇒ \dfrac{k - 5}{h - 3} = - 1 ⇒ k + h = 8 . . . \left(1\right)$

Mid-point of $P{P}^{'}$ lies on the given line,

$⇒ \left(\dfrac{h + 3}{2}\right) - \left(\dfrac{k + 5}{2}\right) + 1 = 0 ⇒ h + 3 - k - 5 + 2 = 0 ⇒ h - k = 0 . . . \left(2\right)$Solving equation \left(1\right)$and \left(2\right)$ we get, $h = 4, k = 4$

Which satisfy the equation ${\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = 4$.