JEE Mathematics 2026 Question with Solution

Given: The circles are $(x + 1)^2 + (y + 4)^2 = r^2$ and $x^2 + y^2 - 4x - 2y - 4 = 0$.

Find: The value of $\alpha\beta$ where the values of $r$ for intersection at two distinct points form the interval $(\alpha, \beta)$.

For the first circle, the center is $C_1(-1,-4)$ and the radius is $r$.

For the second circle, complete the square:

$$x^2 - 4x + y^2 - 2y = 4$$ $$(x - 2)^2 + (y - 1)^2 = 9$$

So its center is $C_2(2,1)$ and its radius is $3$.

Now find the distance between the centers:

$$d = \sqrt{(2+1)^2 + (1+4)^2} = \sqrt{3^2 + 5^2} = \sqrt{34}$$

For two circles to intersect at two distinct points, the condition is:

$$|r-3| < \sqrt{34} < r+3$$

From

$$\sqrt{34} < r+3$$

we get

$$r > \sqrt{34} - 3$$

From

$$|r-3| < \sqrt{34}$$

we get

$$-\sqrt{34} < r-3 < \sqrt{34}$$

So,

$$r < 3 + \sqrt{34}$$

Hence,

$$(\alpha, \beta) = (\sqrt{34} - 3, \sqrt{34} + 3)$$

Therefore,

$$\alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3) = 34 - 9 = 25$$

So the correct option is A.

Given: Two circles with centers and radii obtained from their standard forms.

Find: The product $\alpha\beta$.

The second circle is first converted to standard form:

$$x^2 + y^2 - 4x - 2y - 4 = 0$$ $$x^2 - 4x + y^2 - 2y = 4$$ $$(x-2)^2 + (y-1)^2 = 9$$

Thus the two circles have:

• center $(-1,-4)$ and radius $r$
• center $(2,1)$ and radius $3$

Distance between centers:

$$d = \sqrt{(2-(-1))^2 + (1-(-4))^2} = \sqrt{3^2 + 5^2} = \sqrt{34}$$

For two distinct intersection points, the distance must satisfy:

$$|r_1-r_2| < d < r_1+r_2$$

Substituting $r_1=r$, $r_2=3$ and $d=\sqrt{34}$:

$$|r-3| < \sqrt{34} < r+3$$

This gives:

$$r > \sqrt{34}-3$$

and

$$r < \sqrt{34}+3$$

Therefore,

$$\alpha = \sqrt{34}-3, \qquad \beta = \sqrt{34}+3$$

Now multiply:

$$\alpha\beta = (\sqrt{34}-3)(\sqrt{34}+3)$$ $$\alpha\beta = 34-9 = 25$$

Therefore, the required value is $25$.