Radius of Circle Passing Through Tangency Points on Conics

To find the radius of the circle passing through the points $P$, $Q$, and $R$, we first determine the coordinates of each point based on the given conditions.

Step 1: Finding the coordinates of point $P$

The equation of the parabola is given by: $y = x^2$

Differentiating with respect to $x$: $\frac{dy}{dx} = 2x$

The slope of the tangent at point $P$ is given to be $4$: $2x = 4 \implies x = 2$

Substituting $x = 2$ into the parabola's equation: $y = 2^2 = 4$

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

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Step 2: Finding the coordinates of point $Q$

The equation of the circle is given by: $x^2 + y^2 = 2$

Differentiating implicitly with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$

The slope of the tangent at point $Q$ is given to be $-1$: $-\frac{x}{y} = -1 \implies x = y$

Since $Q$ lies in the first quadrant ($x > 0, y > 0$): $x^2 + x^2 = 2 \implies 2x^2 = 2 \implies x = 1$

Thus, $y = 1$, and the coordinates of point $Q$ are: $Q = (1, 1)$

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Step 3: Finding the coordinates of point $R$

The equation of the ellipse is given by: $x^2 + 4y^2 = 8$

Differentiating implicitly with respect to $x$: $2x + 8y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{4y}$

The slope of the tangent at point $R$ is given to be $-\frac{1}{2}$: $-\frac{x}{4y} = -\frac{1}{2} \implies x = 2y$

Since $R$ lies in the first quadrant ($x > 0, y > 0$): $(2y)^2 + 4y^2 = 8 \implies 4y^2 + 4y^2 = 8 \implies 8y^2 = 8 \implies y = 1$

Substituting $y = 1$ to find $x$: $x = 2(1) = 2$

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

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Step 4: Finding the radius of the circle passing through $P$, $Q$, and $R$

We have the three points: $P(2, 4), \quad Q(1, 1), \quad R(2, 1)$

Observe the properties of $\triangle PQR$:

• The line segment $PR$ lies along the vertical line $x = 2$.
• The line segment $QR$ lies along the horizontal line $y = 1$.

Since $PR$ is parallel to the y-axis and $QR$ is parallel to the x-axis, the angle at vertex $R$ is a right angle: $\angle PRQ = 90^\circ$

In a right-angled triangle, the circumcircle's center is the midpoint of the hypotenuse $PQ$, and its radius $r$ is half the length of the hypotenuse $PQ$.

The length of the hypotenuse $PQ$ is: $PQ = \sqrt{(2 - 1)^2 + (4 - 1)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$

Therefore, the radius $r$ of the circle passing through $P$, $Q$, and $R$ is: $r = \frac{PQ}{2} = \frac{\sqrt{10}}{2} = \sqrt{\frac{5}{2}}$

Correct Option: C