Two tangent lines l _ 1 and l _ 2 are drawn from the point (2, 0)...

Given, from point $\left(2, 0\right)$ tangents are drawn on $2{y}^{2} = - x$, these tangents also touches the circle ${\left(x - 5\right)}^{2} + {y}^{2} = {r}^{2}$, then value of $17{r}^{2} is:</p><p>Now equation of tangents when point is outside will be$y = mx - \dfrac{1}{8m}$, now \left(2, 0\right)$ will lie here so $m = \pm \dfrac{1}{4}$, putting the value of $m$ we get,

$x - 4y = 2 and 4y + x = 2$

Using distance of centre from tangent equal to radius, we get

\left|\dfrac{5 - 0 - 2}{\sqrt{17}}\right| = r$or \left|\dfrac{5 + 0 - 2}{\sqrt{17}}\right| = r$

$∴ {r}^{2} = \dfrac{9}{17}$

$⇒ 17{r}^{2} = 9$