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Maths Question 10 – JEE-MAIN 2026

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Let P be a moving point on the circle x2+y2−6x−8y+21=0. Then, the maximum distance of P from the vertex of the parabola x2+6x+y+13=0 is equal to:

The maximum distance from an external point to a point on a circle occurs along the line connecting the external point to the center of the circle, extended outwards.

Step 1: Determine the center and radius of the circle✦ Active

The equation of the circle is x2+y2−6x−8y+21=0. We complete the square to find its standard form:

(x2−6x+9)+(y2−8y+16)=−21+9+16 (x−3)2+(y−4)2=4

The center of the circle is C(3,4) and the radius is R=4=2.

Step 2: Determine the vertex of the parabola○ Expand

The equation of the parabola is x2+6x+y+13=0. We rearrange it to the standard form for a parabola:

y=−x2−6x−13 y=−(x2+6x+9)+9−13 y=−(x+3)2−4 (x+3)2=−(y+4)

The vertex of the parabola is V(−3,−4).

Step 3: Calculate the maximum distance○ Expand

First, calculate the distance between the center of the circle C(3,4) and the vertex of the parabola V(−3,−4) using the distance formula:

CV=(3−(−3))2+(4−(−4))2 CV=(6)2+(8)2=36+64=100=10

The maximum distance of a point P on the circle from the vertex V is the sum of the distance CV and the radius R:

Maximum distance=CV+R=10+2=12
💡 Teacher's Secret Hint

Remember that the maximum distance from an external point to a circle is found by adding the radius to the distance between the external point and the circle's center.

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