Let $P$ be the point on the parabola $y=x^2$ such that the slope of the tangent to the parabola at the point $P$ is $4$. Let $Q$ be the point in the first quadrant lying on the circle $x^2+y^2=2$ such that the slope of the tangent to the circle at the point $Q$ is $-1$. Let $R$ be the point in the first quadrant lying on the ellipse $x^2+4y^2=8$ such that the slope of the tangent to the ellipse at the point $R$ is $-\frac{1}{2}$. Then the radius of the circle passing through the points $P,Q$ and $R$ is:

The correct option is $\sqrt{\frac{5}{2}}$.

To find the radius of the circle passing through the points $P$, $Q$, and $R$, we first need to determine the exact coordinates of these three points.

Step 1: Find the coordinates of point $P$
Point $P$ lies on the parabola $y=x^2$.
Differentiating with respect to $x$ gives the slope of the tangent:

$\frac{dy}{dx}=2x$

We are given that the slope of the tangent at point $P(x_1,y_1)$ is $4$:

$2x_1=4\Rightarrow x_1=2$

Substituting $x_1=2$ back into the parabola's equation:

$y_1=2^2=4$

Thus, the coordinates of point $P$ are $P(2,4)$.

Step 2: Find the coordinates of point $Q$
Point $Q$ lies in the first quadrant on the circle $x^2+y^2=2$.
Differentiating implicitly with respect to $x$:

$2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$

We are given that the slope at $Q(x_2,y_2)$ is $-1$:

$-\frac{x_2}{y_2}=-1\Rightarrow x_2=y_2$

Substituting $x_2=y_2$ into the circle's equation:

$x_2^2+x_2^2=2\Rightarrow 2x_2^2=2\Rightarrow x_2^2=1$

Since point $Q$ lies in the first quadrant, $x_2>0$ and $y_2$. Therefore:

$x_2=1, y_2=1$

Thus, the coordinates of point $Q$ are $Q(1,1)$.

Step 3: Find the coordinates of point $R$
Point $R$ lies in the first quadrant on the ellipse $x^2+4y^2=8$.
Differentiating implicitly with respect to $x$:

$2x+8y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{4y}$

We are given that the slope at $R(x_3,y_3)$ is $-\frac{1}{2}$:

$-\frac{x_3}{4y_3}=-\frac{1}{2}\Rightarrow x_3=2y_3$

Substituting $x_3=2y_3$ into the ellipse equation:

$(2y_3{)}^2+4y_3^2=8\Rightarrow 4y_3^2+4y_3^2=8\Rightarrow 8y_3^2=8\Rightarrow y_3^2=1$

Since point $R$ lies in the first quadrant, $y_3>0$:

$y_3=1, x_3=2\left(1\right)=2$

Thus, the coordinates of point $R$ are $R(2,1)$.

Step 4: Find the radius of the circle passing through $P(2,4)$, $Q(1,1)$, and $R(2,1)$
Notice the points: $P(2,4)$, $Q(1,1)$, and $R(2,1)$.
Let us analyze the side slopes of triangle $PQR$:
The line segment $PR$ lies on the vertical line $x=2$.
The line segment $QR$ lies on the horizontal line $y=1$.

Since $PR$ is vertical and $QR$ is horizontal, the angle between them at vertex $R$ is a right angle:

$\angle PRQ={90}^{\circ }$

Therefore, $△PQR$ is a right-angled triangle with hypotenuse $PQ$.
The circle passing through all three vertices of a right-angled triangle has the hypotenuse $PQ$ as its diameter.

Step 5: Calculate the radius
The length of hypotenuse $PQ$ using the distance formula between $P(2,4)$ and $Q(1,1)$ is:

$\text{Diameter}=PQ=\sqrt{(2-1{)}^2+(4-1{)}^2}$

$PQ=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}$

The radius $r$ is half of the diameter $PQ$:

$r=\frac{\sqrt{10}}{2}=\sqrt{\frac{10}{4}}=\sqrt{\frac{5}{2}}$

Thus, the radius of the circumcircle is $\sqrt{\frac{5}{2}}$.