If the normals to curve $y = x^2$ at the points $P, Q$ and $R$ pass through the point $\left(0, \frac{3}{2}\right)$, then the radius of the circle circumscribing $\triangle PQR$ is ______.
Step-by-Step Solution
Key Concept: Find points on parabola y = x² where normals pass through a given point by setting up the normal equation y - t² = -1/(2t)(x - t) and requiring it to satisfy the condition. The three parameter values yield three points forming a triangle whose circumradius equals half the hypotenuse (for right triangles) or can be found using the circumradius formula R = abc/(4K).
The equation of the normal is $y - t^2 = \frac{1}{2t}(x - t)$ and passes through $\left(0, \frac{3}{2}\right)$. Setting up the equation gives $t = 0$ or $t = \pm 1$, yielding three points $P$, $Q$, $R$ as $(0,0)$, $(1,1)$, and $(-1,1)$. These form a right triangle with the circumcircle having radius $1$.
Correct Answer: 1