Coordinate Geometry – Circles – JEE Main 2 Apr 2025 Shift 1

Solution:

The lines are $L_1: x+y-3=0$ and $L_2: x-y-3=0$.
The center of a circle touching these lines must lie on their angle bisectors.
Bisectors: $\frac{x+y-3}{\sqrt{2}} = \pm \frac{x-y-3}{\sqrt{2}} \Rightarrow y=0 \text{ or } x=3$.

Let the center be $(h, k)$. Since the circle passes through $(-9, 4)$, let’s check which bisector is feasible.
Assuming the center lies on $y=0$, let center be $C(h, 0)$.
The radius $r$ is the perpendicular distance from $C$ to either line:
$$ r = \left| \frac{h+0-3}{\sqrt{2}} \right| = \frac{|h-3|}{\sqrt{2}} $$

Since the circle passes through $P(-9, 4)$, the distance $CP$ equals $r$:
$$ (h+9)^2 + (0-4)^2 = r^2 = \frac{(h-3)^2}{2} $$
$$ 2(h^2 + 18h + 81 + 16) = h^2 – 6h + 9 $$
$$ 2h^2 + 36h + 194 = h^2 – 6h + 9 $$
$$ h^2 + 42h + 185 = 0 $$

The roots $h_1, h_2$ give the centers of the two circles.
Solving for $h$:
$$ h = \frac{-42 \pm \sqrt{1764 – 740}}{2} = \frac{-42 \pm \sqrt{1024}}{2} = \frac{-42 \pm 32}{2} $$
$$ h_1 = -5, \quad h_2 = -37 $$

Calculate the squares of the radii:
For $h_1 = -5$:
$$ r_1^2 = \frac{(-5-3)^2}{2} = \frac{64}{2} = 32 $$
For $h_2 = -37$:
$$ r_2^2 = \frac{(-37-3)^2}{2} = \frac{1600}{2} = 800 $$

The absolute difference:
$$ |r_1^2 – r_2^2| = |32 – 800| = 768 $$

Ans. (768)