If the common tangent to the parabolas, y ^ 2 = 4xand x ^ 2 = 4y...

Given: ${y}^{2} = 4x \text{ and } {x}^{2} = 4y$

Comparing with standard equations, ${y}^{2} = 4{a}_{1}x \text{ and } {x}^{2} = 4{a}_{2}y$, ${a}_{1} = {a}_{2} = 1$

Therefore tangent of ${y}^{2} = 4x$ is $y = mx + \dfrac{1}{m}$ which is also tangent to ${x}^{2} = 4y$, whose equation would be $y = mx - {m}^{2}$

$∴ \dfrac{1}{m} = - {m}^{2} ⇒ m = - 1$

$∴$ common tangent is $y = - x - 1,$ which touches ${x}^{2} + {y}^{2} = {c}^{2}$

$∴ ( - 1{)}^{2} = {c}^{2} ( 1 + ( - 1{)}^{2} ) ⇒ 1 = {c}^{2} \cdot ( 1 + 1 ) ⇒ {c}^{2} = \dfrac{1}{2}$