JEE Main 2025 Apr 8 Shift 2, Mathematics Q25: Involving Two or More Circles

JEE Main2025Apr 8, Shift 2Mathematics

Let $r$ be the radius of the circle, which touches x-axis at point $(a, 0)$ , $a < 0$ and the parabola $y^2 = 9x$ at the point $(4, 6)$. Then $r$ is equal to ______

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Solution

Since the circle touches the x-axis at $(a, 0)$ , its centre lies directly above this point at $C = (a, r)$ with radius $r$ .

The point $(4, 6)$ lies on the circle: $(4 - a)^{2} + (6 - r)^{2} = r^{2}$ , expanding to $a^{2} - 8 a - 12 r + 52 = 0$ .

Compute the tangent to $y^{2} = 9 x$ at $(4, 6)$ using $y y_{1} = \frac{9}{2} (x + x_{1})$ : $6 y = \frac{9}{2} (x + 4)$ , i.e. $3 x - 4 y + 12 = 0$ .

This tangent line must also be tangent to the circle, so the perpendicular distance from centre $(a, r)$ equals $r$ : $\frac{∣3 a - 4 r + 12∣}{5} = r$ , giving $3 a - 4 r + 12 = \pm 5 r$ .

Case 1: $3 a + 12 = 9 r$ , so $a = 3 r - 4$ .

Substituting into the circle equation yields $a = 6$ , which contradicts $a < 0$.

Case 2: $3 a + 12 = - r$ , so $a = - \frac{r + 12}{3}$ . Substituting into $a^{2} - 8 a - 12 r + 52 = 0$ and solving gives $a = - 14$ and $r = 30$ .

Therefore $r = 30$ .

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q25: Involving Two or More Circles

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