JEE Mathematics 2025 Question with Solution

Given: The points $(4, 6)$, $(-1, 5)$, $(0, 0)$ and $(k, 3k)$ lie on the same circle.

Find: The value of $10k + r^2$.

Let the equation of the circle be

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Since $(0, 0)$ lies on the circle, we get

$$c = 0$$

Substitute $(4, 6)$:

$$16 + 36 + 8g + 12f = 0$$ $$8g + 12f = -52$$ $$2g + 3f = -13 \quad \text{...(i)}$$

Substitute $(-1, 5)$:

$$1 + 25 - 2g + 10f = 0$$ $$-2g + 10f = -26$$ $$-g + 5f = -13 \quad \text{...(ii)}$$

Add $\text{(i)}$ and $2 \times \text{(ii)}$:

$$2g + 3f + 2(-g + 5f) = -13 + 2(-13)$$ $$13f = -39$$ $$f = -3$$

Substitute $f = -3$ in $\text{(ii)}$:

$$-g + 5(-3) = -13$$ $$-g - 15 = -13$$ $$-g = 2$$ $$g = -2$$

Therefore, the circle is

$$x^2 + y^2 - 4x - 6y = 0$$

Its center is $(2, 3)$ and radius is

$$r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{13}$$

So,

$$r^2 = 13$$

Now $(k, 3k)$ also lies on the circle, so substitute in the equation:

$$k^2 + (3k)^2 - 4k - 6(3k) = 0$$ $$k^2 + 9k^2 - 4k - 18k = 0$$ $$10k^2 - 22k = 0$$ $$2k(5k - 11) = 0$$

Since the four points are distinct, $k \ne 0$. Hence,

$$5k - 11 = 0$$ $$k = \frac{11}{5}$$

Now compute:

$$10k + r^2 = 10 \cdot \frac{11}{5} + 13 = 22 + 13 = 35$$

Therefore, the correct option is D.

Given: Three fixed points $(4, 6)$, $(-1, 5)$ and $(0, 0)$ determine the circle, and the fourth point $(k, 3k)$ also lies on it.

Find: $10k + r^2$.

From the general circle equation,

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Using $(0, 0)$ gives $c = 0$.

Using $(4, 6)$ and $(-1, 5)$ gives the two linear equations

$$2g + 3f = -13$$ $$-g + 5f = -13$$

Solving them gives

$$f = -3, \qquad g = -2$$

Hence the circle is

$$x^2 + y^2 - 4x - 6y = 0$$

Comparing with standard form, the center is $(2, 3)$.

Now radius squared is the distance squared from the center to $(0, 0)$:

$$r^2 = (2-0)^2 + (3-0)^2 = 4 + 9 = 13$$

For the point $(k, 3k)$ to lie on the same circle,

$$(k-2)^2 + (3k-3)^2 = 13$$

Expand:

$$k^2 - 4k + 4 + 9k^2 - 18k + 9 = 13$$ $$10k^2 - 22k + 13 = 13$$ $$10k^2 - 22k = 0$$ $$2k(5k-11) = 0$$

Since $k = 0$ would repeat the point $(0, 0)$, we must have

$$k = \frac{11}{5}$$

Therefore,

$$10k + r^2 = 10 \cdot \frac{11}{5} + 13 = 35$$

So the required value is $35$.