JEE Mathematics 2025 Question with Solution

Given: The circle touches the $x$-axis at $(a,0)$ with $a<0$ and touches the parabola $y^2=9x$ at $(4,6)$.

Find: The radius $r$ of the circle.

For the parabola, the tangent at $(4,6)$ is taken as

$$3x - 4y + 12 = 0$$

The circle through the point of contact $(4,6)$ and tangent to this line is written as

$$(x - 4)^2 + (y - 6)^2 + \lambda(3x - 4y + 12) = 0$$

Expanding,

$$x^2 + y^2 + (3\lambda - 8)x + (-12 - 4\lambda)y + 52 + 12\lambda = 0$$

From the comparison used in the solution,

$$\left(\frac{3\lambda - 8}{2}\right)^2 = 52 + 12\lambda$$

So,

$$9\lambda^2 + 64 - 48\lambda = 208 + 48\lambda$$

which gives

$$9\lambda^2 - 96\lambda - 144 = 0$$

Solving,

$$\lambda = 12, -\frac{2}{3}$$

Also,

$$f = -(2\lambda + 6)$$

so

$$f = -30, -\frac{14}{3}$$

The radius is

$$r = \sqrt{g^2 + f^2 - c} = |f| = |-(2\lambda + 6)|$$

Since the center lies in the second quadrant,

$$3\lambda - 8 > 0 \Rightarrow \lambda > \frac{8}{3}$$

Hence, $\lambda = 12$ and therefore

$$r = 30$$

Therefore, the radius is $30$.

The extracted hint says to use the fact that the point of tangency must satisfy both the parabola and the circle, together with the geometric relation involving the tangent line.

After forming the circle using the tangent at $(4,6)$, the parameter values obtained are

$$\lambda = 12, -\frac{2}{3}$$

Then

$$r = |-(2\lambda + 6)|$$

The condition that the center lies in the second quadrant gives

$$3\lambda - 8 > 0$$

so only

$$\lambda = 12$$

is admissible. Therefore,

$$r = 30$$

So the required numerical value is $30$.