Consider the region R=\ (x, y) ∈ ℝ × ℝ: x ≥ 0. and .y^ 2 ≤ 4-x\ ....

equation of normal at $P$ is $y - \beta = 2\beta\left(x - 4 + {\beta}^{2}\right)</p><p>Passing through$( r, 0 )$</p><p>$⇒ - 1 = 2 \left(r - 4 + {\beta}^{2}\right) . . . $\left(i\right)</p><p>or$\beta = 0 . . . \left(ii\right)$</p><p>Also${r}^{2} = {\left(r + {\beta}^{2} - 4\right)}^{2} + {\beta}^{2} . . . \left(iii\right)

from \left(i\right)$and \left(iii\right)$

$r = \dfrac{3}{2}, - \dfrac{5}{2}$

so radius$of circle is1 . 5$

and $\alpha = 2$

from \left(ii\right)$and \left(iii\right)$

$r = 2$ but in this case circle intersect at three points