JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Circle

The absolute difference between the squares of the radii of the two circles passing through the point

(- 9, 4)

and touching the lines

x + y = 3

and

x - y = 3

, is equal to ______.

Answer: 768

Solution

Since the two lines

x + y = 3

and

x - y = 3

are perpendicular and intersect at

(3, 0)

, any circle tangent to both lines must have its center on the angle bisectors. By symmetry, the centers lie on the

x

-axis at points

(a, 0)

.

For a circle centered at

(a, 0)

to be tangent to both lines, the radius is the perpendicular distance from

(a, 0)

to either line:

r = \frac{a - 3}{2}

The circle equation is

(x - a)^{2} + y^{2} = \frac{( a - 3 ) ^{2}}{2}

.

Since the circle passes through

(- 9, 4)

(- 9 - a)^{2} + 16 = \frac{( a - 3 ) ^{2}}{2}

2 (a^{2} + 18 a + 81 + 16) = a^{2} - 6 a + 9

2 a^{2} + 36 a + 194 = a^{2} - 6 a + 9

a^{2} + 42 a + 185 = 0

(a + 37) (a + 5) = 0

a = - 37

a = - 5

.

The radii are:

r_{1} = \frac{∣ - 37 - 3∣}{2} = \frac{40}{2} = 202

r_{2} = \frac{∣ - 5 - 3∣}{2} = \frac{8}{2} = 42

Therefore:

∣ r_{1}^{2} - r_{2}^{2} ∣ = ∣800 - 32∣ = 768

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Integer answer