Maths Q11 – JEE-MAIN 2025

If the four distinct points

(4, 6)

(- 1, 5)

(0, 0)

and

(k, 3 k)

lie on a circle of radius

r

, then

10 k + r^{2}

is equal to

Clue 1 of 3: General Equation of a Circle

Start by using the general equation of a circle, $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , and the fact that one of the given points is the origin to simplify it.

🧠 Full Solution Path

Step 1: Determine the equation of the circle✦ Active

The general equation of a circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Since the point $(0, 0)$ lies on the circle, substituting it gives $c = 0$ . Thus, the equation simplifies to $x^{2} + y^{2} + 2 g x + 2 f y = 0$ .

Substitute the point $(4, 6)$ : $4^{2} + 6^{2} + 2 g (4) + 2 f (6) = 0 ⟹ 16 + 36 + 8 g + 12 f = 0 ⟹ 8 g + 12 f = - 52 ⟹ 2 g + 3 f = - 13 (1)$ .

Substitute the point $(- 1, 5)$ : $(- 1)^{2} + 5^{2} + 2 g (- 1) + 2 f (5) = 0 ⟹ 1 + 25 - 2 g + 10 f = 0 ⟹ - 2 g + 10 f = - 26 ⟹ - g + 5 f = - 13 (2)$ .

Solve the system of equations (1) and (2). From (2), $g = 5 f + 13$ . Substitute this into (1): $2 (5 f + 13) + 3 f = - 13 ⟹ 10 f + 26 + 3 f = - 13 ⟹ 13 f = - 39 ⟹ f = - 3$ . Substitute $f = - 3$ back into $g = 5 f + 13$ : $g = 5 (- 3) + 13 = - 15 + 13 = - 2$ .

The equation of the circle is $x^{2} + y^{2} - 4 x - 6 y = 0$ .

Step 2: Calculate

r^{2}

and find

k

○ Expand

The radius squared is $r^{2} = g^{2} + f^{2} - c$ . Since $c = 0$ , $r^{2} = (- 2)^{2} + (- 3)^{2} = 4 + 9 = 13$ .

The fourth point $(k, 3 k)$ lies on the circle. Substitute it into the circle's equation: $k^{2} + (3 k)^{2} - 4 k - 6 (3 k) = 0$ .

$k^{2} + 9 k^{2} - 4 k - 18 k = 0 ⟹ 10 k^{2} - 22 k = 0$ .

Factor out $2 k$ : $2 k (5 k - 11) = 0$ . Since the four points are distinct, $k \neq 0$ (otherwise $(k, 3 k)$ would be $(0, 0)$ ). Therefore, $5 k - 11 = 0 ⟹ k = \frac{11}{5}$ .

Step 3: Calculate the final expression○ Expand

We need to find the value of $10 k + r^{2}$ .

Substitute the values of $k$ and $r^{2}$ : $10 (\frac{11}{5}) + 13 = 2 \times 11 + 13 = 22 + 13 = 35$ .

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Maths Question 11 – JEE-MAIN 2025

Clue 1 of 3: General Equation of a Circle

🧠 Full Solution Path