Maths Q10 – JEE-MAIN 2026

Let P be a moving point on the circle

x^{2} + y^{2} - 6 x - 8 y + 21 = 0

. Then, the maximum distance of P from the vertex of the parabola

x^{2} + 6 x + y + 13 = 0

is equal to:

Clue 1 of 3: Geometric Interpretation

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.

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● Video Walkthrough

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

The equation of the circle is $x^{2} + y^{2} - 6 x - 8 y + 21 = 0$ . We complete the square to find its standard form:

(x^{2} - 6 x + 9) + (y^{2} - 8 y + 16) = - 21 + 9 + 16

(x - 3)^{2} + (y - 4)^{2} = 4

The center of the circle is $C (3, 4)$ and the radius is $R = \sqrt{4} = 2$ .

Step 2: Determine the vertex of the parabola○ Expand

The equation of the parabola is $x^{2} + 6 x + y + 13 = 0$ . We rearrange it to the standard form for a parabola:

y = - x^{2} - 6 x - 13

y = - (x^{2} + 6 x + 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:

C V = \sqrt{(3 - (- 3))^{2} + (4 - (- 4))^{2}}

C V = \sqrt{(6)^{2} + (8)^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10

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

Maximum distance = C V + 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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Maths Question 10 – JEE-MAIN 2026

Clue 1 of 3: Geometric Interpretation

🧠 Full Solution Path