MHTCET 19 Apr 2026 Shift 2 Paper

Concept:A circle touching a parabola at a point has its centre on the normal to the parabola at that point.Also, the centre is equidistant from any two points on the circle.Explanation:The circle passes through the two points $(0,1)$ and $(1,1)$.The perpendicular bisector of the chord joining these points must pass through the centre.Since the chord is horizontal, its perpendicular bisector is the vertical line $x = \frac{1}{2}$.Thus, the $x$-coordinate of the centre is $\frac{1}{2}$.Now, for the parabola $y = x^2$, the slope of the tangent is given by $\frac{dy}{dx} = 2x$.At the point of contact $(1,1)$, the slope of the tangent is $2(1) = 2$.Therefore, the slope of the normal at $(1,1)$ is $-\frac{1}{2}$.The equation of the normal is:$y - 1 = -\frac{1}{2}(x - 1)$.Since the centre lies on this normal, substitute $x = \frac{1}{2}$:$y - 1 = -\frac{1}{2}\left(\frac{1}{2} - 1\right) = -\frac{1}{2}\left(-\frac{1}{2}\right) = \frac{1}{4}$.So, $y = 1 + \frac{1}{4} = \frac{5}{4}$.Hence, the centre of the circle is $\left(\frac{1}{2}, \frac{5}{4}\right)$.Answer:The centre of the circle is $\left(\frac{1}{2}, \frac{5}{4}\right)$, which corresponds to Option C.