Maths Q10 – JEE-MAIN 2026

Let C be a circle having centre in the first quadrant and touching the x-axis at a distance of 3 units from the origin. If the circle C has an intercept of length

6 \sqrt{3}

on y-axis, then the length of the chord of the circle C on the line

x - y = 3

is:

Clue 1 of 3: Circle Properties

Understand how the center and radius of a circle relate to its tangency with axes and intercepts.

🧠 Full Solution Path

Step 1: Determine the Circle's Equation✦ Active

The circle's center is in the first quadrant and touches the x-axis at $(3, 0)$ . This implies the center is $(3, r)$ and the radius is $r$ . The equation of the circle is $(x - 3)^{2} + (y - r)^{2} = r^{2}$ . The y-intercept length is $6 \sqrt{3}$ . Setting $x = 0$ gives $9 + (y - r)^{2} = r^{2} \Rightarrow y^{2} - 2 r y + 9 = 0$ . The length of the y-intercept is $\sqrt{(- 2 r)^{2} - 4 (1) (9)} = \sqrt{4 r^{2} - 36}$ . Equating this to $6 \sqrt{3}$ :

\sqrt{4 r^{2} - 36} = 6 \sqrt{3} \Rightarrow 4 r^{2} - 36 = 108 \Rightarrow 4 r^{2} = 144 \Rightarrow r^{2} = 36 \Rightarrow r = 6

Thus, the center is $(3, 6)$ and the radius is $R = 6$ . The circle's equation is $(x - 3)^{2} + (y - 6)^{2} = 36$ .

Step 2: Calculate Perpendicular Distance to the Line○ Expand

The line is $x - y - 3 = 0$ . The center of the circle is $(3, 6)$ . The perpendicular distance $d$ from the center to the line is calculated using the formula $d = \frac{| A x_{0} + B y_{0} + C |}{\sqrt{A^{2} + B^{2}}}$ :

d = \frac{| 1 (3) + (- 1) (6) - 3 |}{\sqrt{1^{2} + (- 1)^{2}}} = \frac{| 3 - 6 - 3 |}{\sqrt{2}} = \frac{| - 6 |}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3 \sqrt{2}

Step 3: Calculate the Length of the Chord○ Expand

Using the relation $R^{2} = d^{2} + (L / 2)^{2}$ , where $R = 6$ is the radius, $d = 3 \sqrt{2}$ is the perpendicular distance, and $L$ is the chord length:

6^{2} = (3 \sqrt{2})^{2} + {(\frac{L}{2})}^{2}

Simplifying the equation:

36 = 18 + \frac{L^{2}}{4} \Rightarrow 18 = \frac{L^{2}}{4} \Rightarrow L^{2} = 72 \Rightarrow L = \sqrt{72} = 6 \sqrt{2}

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Maths Question 10 – JEE-MAIN 2026

Clue 1 of 3: Circle Properties

🧠 Full Solution Path