Maths Q23 – JEE-MAIN 2025

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 _______

Enter Calculated Numerical Value:

Clue 1 of 3: Geometric Properties

Understand that a circle touching the x-axis at $(a, 0)$ with radius $r$ has its center at $(a, r)$ or $(a, - r)$ . Given the point of tangency on the parabola, determine the correct center.

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Step 1: Formulate Circle Equation and Point Condition✦ Active

Since the circle touches the $x$ -axis at $(a, 0)$ and the point $(4, 6)$ (where $y > 0$ ) lies on the circle, the center of the circle must be $(a, r)$ with radius $r$ . The equation of the circle is $(x - a)^{2} + (y - r)^{2} = r^{2}$ . As the point $(4, 6)$ lies on the circle, we substitute it into the equation:

(4 - a)^{2} + (6 - r)^{2} = r^{2}

(16 - 8 a + a^{2}) + (36 - 12 r + r^{2}) = r^{2}

a^{2} - 8 a - 12 r + 52 = 0 (Equation 1)

Step 2: Apply Tangency Condition (Normal to Parabola)○ Expand

The circle and the parabola $y^{2} = 9 x$ touch at $(4, 6)$ . This implies that the normal to the parabola at $(4, 6)$ passes through the center of the circle $(a, r)$ . First, find the slope of the tangent to the parabola:

y^{2} = 9 x 2 y \frac{d y}{d x} = 9 \frac{d y}{d x} = \frac{9}{2 y}

At $(4, 6)$ , the slope of the tangent is $m_{t} = \frac{9}{2 \times 6} = \frac{3}{4}$ . The slope of the normal is $m_{n} = - \frac{1}{m_{t}} = - \frac{4}{3}$ . The equation of the normal passing through $(4, 6)$ is $y - 6 = - \frac{4}{3} (x - 4)$ . Since the center $(a, r)$ lies on this normal:

r - 6 = - \frac{4}{3} (a - 4)

3 (r - 6) = - 4 (a - 4)

3 r - 18 = - 4 a + 16

4 a + 3 r = 34 (Equation 2)

💡 Teacher's Secret Hint

Remember that the radius of a circle is perpendicular to the tangent at the point of contact, meaning it lies along the normal.

Step 3: Solve the System of Equations for r○ Expand

From Equation 2, express $a$ in terms of $r$ : $a = \frac{34 - 3 r}{4}$ . Substitute this into Equation 1:

{(\frac{34 - 3 r}{4})}^{2} - 8 (\frac{34 - 3 r}{4}) - 12 r + 52 = 0

\frac{(34 - 3 r)^{2}}{16} - 2 (34 - 3 r) - 12 r + 52 = 0

Multiply by 16 to clear the denominator:

(34 - 3 r)^{2} - 32 (34 - 3 r) - 192 r + 832 = 0

1156 - 204 r + 9 r^{2} - 1088 + 96 r - 192 r + 832 = 0

9 r^{2} - 300 r + 900 = 0

3 r^{2} - 100 r + 300 = 0

Solve this quadratic equation for $r$ using the quadratic formula $r = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

r = \frac{100 \pm \sqrt{(- 100)^{2} - 4 (3) (300)}}{2 (3)}

r = \frac{100 \pm \sqrt{10000 - 3600}}{6}

r = \frac{100 \pm \sqrt{6400}}{6}

r = \frac{100 \pm 80}{6}

This gives two possible values for $r$ : $r_{1} = \frac{100 + 80}{6} = \frac{180}{6} = 30$ and $r_{2} = \frac{100 - 80}{6} = \frac{20}{6} = \frac{10}{3}$ . Now, check the condition $a < 0$ :

For $r = 30$ : $a = \frac{34 - 3 (30)}{4} = \frac{34 - 90}{4} = \frac{- 56}{4} = - 14$ . This satisfies $a < 0$ .

For $r = \frac{10}{3}$ : $a = \frac{34 - 3 (\frac{10}{3})}{4} = \frac{34 - 10}{4} = \frac{24}{4} = 6$ . This does not satisfy $a < 0$ .

Therefore, the only valid radius is $r = 30$ .

💡 Teacher's Secret Hint

Always verify the solutions against all given conditions, especially inequalities like $a < 0$ .

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Maths Question 23 – JEE-MAIN 2025

Clue 1 of 3: Geometric Properties

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