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Maths Question 10 – JEE-MAIN 2026

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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 63 on y-axis, then the length of the chord of the circle C on the line x−y=3 is:

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

● Video Walkthrough
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=r2. The y-intercept length is 63. Setting x=0 gives 9+(y−r)2=r2⇒y2−2ry+9=0. The length of the y-intercept is (−2r)2−4(1)(9)=4r2−36. Equating this to 63:

4r2−36=63⇒4r2−36=108⇒4r2=144⇒r2=36⇒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=|Ax0+By0+C|A2+B2:

d=|1(3)+(−1)(6)−3|12+(−1)2=|3−6−3|2=|−6|2=62=32
Step 3: Calculate the Length of the Chord○ Expand

Using the relation R2=d2+(L/2)2, where R=6 is the radius, d=32 is the perpendicular distance, and L is the chord length:

62=(32)2+(L2)2

Simplifying the equation:

36=18+L24⇒18=L24⇒L2=72⇒L=72=62
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