The absolute difference between the squares of the radii of the t… — Solution | JEE Mathematics

MCQMediumJEE 2025Circle Equation & Properties

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:

A
$768$
B
$550$
C
$860$
D
$999$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Two circles pass through the point $P(-9, 4)$ and touch the lines $x + y = 3$ and $x - y = 3$ .

Find: The absolute difference between the squares of their radii.

The center of a circle tangent to both intersecting lines must lie on an angle bisector of the two lines.

For $x + y - 3 = 0$ and $x - y - 3 = 0$ , the angle bisectors are obtained from

\frac{x + y - 3}{\sqrt{2}} = \pm \frac{x - y - 3}{\sqrt{2}}

So the two bisectors are $y = 0$ and $x = 3$ .

Take the first case where the center is on $y = 0$ . Let the center be $C(h, 0)$ .

Since the circle touches the line $x + y - 3 = 0$ , its radius is the perpendicular distance of $C(h,0)$ from the line:

r = \frac{|h - 3|}{\sqrt{2}}

Hence,

r^2 = \frac{(h - 3)^2}{2}

Because the circle passes through $P(-9,4)$ , we have

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

Multiplying by $2$ ,

2(h^2 + 18h + 97) = h^2 - 6h + 9

So,

h^2 + 42h + 185 = 0

Solving,

h = \frac{-42 \pm \sqrt{1024}}{2} = \frac{-42 \pm 32}{2}

Thus,

h = -5 \quad \text{or} \quad h = -37

Casewise Derivation

For $h = -5$ ,

r_1^2 = \frac{(-5 - 3)^2}{2} = \frac{64}{2} = 32

For $h = -37$ ,

r_2^2 = \frac{(-37 - 3)^2}{2} = \frac{1600}{2} = 800

Reject the Other Bisector Quickly

Now check the second bisector $x = 3$ . Let the center be $C(3,k)$ .

Then the radius is

r = \frac{|k|}{\sqrt{2}}

Using the condition that the circle passes through $P(-9,4)$ ,

144 + (k - 4)^2 = \frac{k^2}{2}

This gives

k^2 - 16k + 320 = 0

Its discriminant is

D = (-16)^2 - 4(1)(320) = -1024

which is negative, so there is no real circle with center on $x = 3$ .

Therefore only the two circles from $y = 0$ are valid, and

|r_2^2 - r_1^2| = |800 - 32| = 768

So the correct option is A.

Common mistakes

Assuming the center can be anywhere in the plane. For a circle tangent to both intersecting lines, the center must lie on an angle bisector. Always find the angle bisectors first.
Using the distance from the center to the point $(-9,4)$ directly as a linear equation instead of equating squared distances. Here, use $CP^2 = r^2$ to avoid sign errors.
Forgetting to test both angle bisectors. One bisector gives the two valid circles, while the other leads to a quadratic with negative discriminant and no real circle.

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