Straight line x – y + 1 = 0 is tangent to the parabola y² = 4x at (1, 1)

$\because$ (x₀, y₀) lying above the line x – y + 1 = 0

$\therefore$        required circle is touches the parabola and the line at (1, 2)

$\therefore$       Family of circle

            (x – 1)² + (y – 2)² + λ(x – y + 1) = 0

            x² + y² + (λ – 2)x – y(λ + 4) + 5 + λ = 0

$\because$         circle is also touching the y-axis.

$\therefore$        f² = c

                      ${\left(\frac{\mathrm{\lambda }+4}{2}\right)}^2=5+\mathrm{\lambda }$

⇒         λ² + 8λ + 16 = 20 + 4λ

⇒         λ² + 4λ + 4 = 8

⇒         (λ + 2)² = 8

$\mathrm{\lambda }=-2\pm 2\sqrt{2}$

$\therefore {\mathrm{x}}^2+{\mathrm{y}}^2+(-4\pm 2\sqrt{2})\mathrm{x}-\mathrm{y}(2\pm 2\sqrt{2})+3\pm 2\sqrt{2}=0$

            (x₀, y₀) lying in first quadrant.

$\therefore$        +ve sign is to be taken

${\mathrm{x}}^2+{\mathrm{y}}^2+(-4+2\sqrt{2})\mathrm{x}-(2+2\sqrt{2})\mathrm{y}+3+2\sqrt{2}=0$

$\mathrm{r}={\mathrm{x}}_0=2-\sqrt{2}=\sqrt{2}(\sqrt{2}-1)$

$=\sqrt{2}\tan \frac{\mathrm{\pi }}{8}\equiv \sqrt{\mathrm{p}}\tan \frac{\mathrm{\pi }}{8}$

$\therefore$      p = 2