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

Let P be the point on the parabola y = x² 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 x² + y² = 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 x² + 4y² = 8 such that the slope of the tangent to the ellipse at the point R is -1/2. Then the radius of the circle passing through the points P, Q and R is $\text{Given: } P \in y=x^2,\ \text{tangent slope } 4$ $Q \in x^2+y^2=2,\ \text{slope } -1;\qquad R \in x^2+4y^2=8,\ \text{slope } -\dfrac{1}{2}$ $\text{To find: radius of the circle through } P, Q, R$ $\text{Plan: find } P, Q, R \text{ from tangent slopes, then the circumcircle of } \triangle PQR$ $y=x^2 \Rightarrow y'=2x=4 \Rightarrow x=2 \Rightarrow P(2,4)$ $x^2+y^2=2 \Rightarrow y'=-\dfrac{x}{y}=-1 \Rightarrow x=y$ $2x^2=2 \Rightarrow Q(1,1)$ $x^2+4y^2=8 \Rightarrow y'=-\dfrac{x}{4y}=-\dfrac{1}{2} \Rightarrow x=2y$ $8y^2=8 \Rightarrow y=1,\ x=2 \Rightarrow R(2,1)$ $\text{Centre } (h,k).\ \ PR \text{ vertical } (x=2):\ (4-k)^2=(1-k)^2 \Rightarrow k=\dfrac{5}{2}$ $QR \text{ horizontal } (y=1):\ (1-h)^2=(2-h)^2 \Rightarrow h=\dfrac{3}{2}$ $\text{Centre} = \left(\dfrac{3}{2},\ \dfrac{5}{2}\right)$ $r^2 = \left(2-\dfrac{3}{2}\right)^2 + \left(4-\dfrac{5}{2}\right)^2 = \dfrac{1}{4}+\dfrac{9}{4}$ $r^2 = \dfrac{5}{2} \Rightarrow r = \sqrt{\dfrac{5}{2}} \Rightarrow (3)$ $\boxed{r = \sqrt{\dfrac{5}{2}}}\quad \text{option (3)}$