JEE Mathematics 2024 Question with Solution

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

Find: The range of $r$ for which they intersect at exactly two distinct points.

For the first circle, the center is $(-1,-2)$ and the radius is $r$.

Rewrite the second circle in standard form:

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

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

The distance between the centers is

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

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

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

Substituting $r_1=r$, $r_2=2$ and $d=5$,

$$|r-2| < 5 < r+2$$

From $5 < r+2$, we get

$$r > 3$$

Also, $|r-2|<5$ gives an upper bound satisfied by

$$r < 7$$

Hence,

$$3 < r < 7$$

Therefore, the correct option is C.

The solution contains a discrepancy because one place marks option A, but the worked solution concludes $$3

$$$$

Given: Two circles with centers $C_1=(-1,-2)$ and $C_2=(2,2)$.

Find: Values of $r$ such that the circles cut each other at two distinct points.

The second equation is simplified by completing squares:

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

So the second circle has radius $2$.

Now compute the distance between centers:

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

For exact two-point intersection:

$$|r-2|<5 \quad \text{and} \quad 5<r+2$$

The second inequality gives

$$r>3$$

The first inequality gives

$$-5$$