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

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Circle

Let

r

be the radius of the circle, which touches x-axis at point (

a, 0

a < 0

and the parabola

y^{2} = 9 x

at the point

(4, 6)

. Then r is equal to ______

Answer: 30

Solution

Touching the x-axis at

(a, 0)

with

a < 0

places the centre at

(a, r)

. The tangent to

y^{2} = 9 x

(4, 6)

6 y = \frac{9}{2} (x + 4)

, that is

3 x - 4 y + 12 = 0

. The circle shares this tangent at

(4, 6)

, so its centre lies along the normal there:

(a, r) = (4 + 3 t, 6 - 4 t)

, with radius

∣ t ∣ 3^{2} + 4^{2} = 5∣ t ∣

. Then

r = 6 - 4 t = 5∣ t ∣

. Taking

t < 0

gives

6 - 4 t = - 5 t

, so

t = - 6

, hence

r = 30

and

a = 4 + 3 (- 6) = - 14 < 0

, which satisfies the requirement. Thus

r = 30

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer

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