Q53 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
A circle passes through the point $(0,1)$ and touches the parabola $y = x^2$ at the point $(1,1)$. The centre of the circle is...
- A$\left(-\dfrac{1}{2}, -\dfrac{5}{2}\right)$
- B$\left(\dfrac{1}{2}, -\dfrac{5}{2}\right)$
- Correct answerC$\left(\dfrac{1}{2}, \dfrac{5}{4}\right)$
- D$\left(-\dfrac{1}{2}, \dfrac{5}{4}\right)$
Solution
The equation of the tangent to the parabola $y = x^2$ at the point $(1,1)$ is given by $T = 0$:
$\dfrac{y+1}{2} = x(1) \Rightarrow 2x - y - 1 = 0$
The equation of the family of circles touching the line $2x - y - 1 = 0$ at $(1,1)$ is:
$(x - 1)^2 + (y - 1)^2 + \lambda(2x - y - 1) = 0$
Since the circle passes through the point $(0,1)$, substituting $x = 0$ and $y = 1$ gives:
$(0 - 1)^2 + (1 - 1)^2 + \lambda(2(0) - 1 - 1) = 0$
$1 - 2\lambda = 0 \Rightarrow \lambda = \dfrac{1}{2}$
Substituting $\lambda = \dfrac{1}{2}$ into the equation of the circle:
$(x - 1)^2 + (y - 1)^2 + \dfrac{1}{2}(2x - y - 1) = 0$
$x^2 - 2x + 1 + y^2 - 2y + 1 + x - \dfrac{y}{2} - \dfrac{1}{2} = 0$
$x^2 + y^2 - x - \dfrac{5}{2}y + \dfrac{3}{2} = 0$
The coordinates of the centre of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ are $(-g, -f)$.
Centre $= \left(\dfrac{1}{2}, \dfrac{5}{4}\right)$
Answer: $\left(\dfrac{1}{2}, \dfrac{5}{4}\right)$
$\dfrac{y+1}{2} = x(1) \Rightarrow 2x - y - 1 = 0$
The equation of the family of circles touching the line $2x - y - 1 = 0$ at $(1,1)$ is:
$(x - 1)^2 + (y - 1)^2 + \lambda(2x - y - 1) = 0$
Since the circle passes through the point $(0,1)$, substituting $x = 0$ and $y = 1$ gives:
$(0 - 1)^2 + (1 - 1)^2 + \lambda(2(0) - 1 - 1) = 0$
$1 - 2\lambda = 0 \Rightarrow \lambda = \dfrac{1}{2}$
Substituting $\lambda = \dfrac{1}{2}$ into the equation of the circle:
$(x - 1)^2 + (y - 1)^2 + \dfrac{1}{2}(2x - y - 1) = 0$
$x^2 - 2x + 1 + y^2 - 2y + 1 + x - \dfrac{y}{2} - \dfrac{1}{2} = 0$
$x^2 + y^2 - x - \dfrac{5}{2}y + \dfrac{3}{2} = 0$
The coordinates of the centre of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ are $(-g, -f)$.
Centre $= \left(\dfrac{1}{2}, \dfrac{5}{4}\right)$
Answer: $\left(\dfrac{1}{2}, \dfrac{5}{4}\right)$
Hence, the correct answer is option: 'C'.