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JEE Mathematics 2025 Question with Solution

Let C1C_1 be the circle in the third quadrant of radius 33, that touches both coordinate axes. Let C2C_2 be the circle with center (1,3)(1, 3) that touches C1C_1 externally at the point (α,β)(\alpha, \beta). If (β−α)2=mn(\beta - \alpha)^2 = \frac{m}{n}, and gcd⁡(m,n)=1\gcd(m, n) = 1, then m+nm + n is equal to:

  • A

    99

  • B

    1313

  • C

    2222

  • D

    3131

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: C1C_1 is in the third quadrant, has radius 33, and touches both coordinate axes. Hence its centre is A(−3,−3)A(-3,-3). Circle C2C_2 has centre B(1,3)B(1,3) and touches C1C_1 externally at P(α,β)P(\alpha,\beta).

Find: m+nm+n if (β−α)2=mn(\beta-\alpha)^2 = \frac{m}{n}.

For C1C_1, the equation is

(x+3)2+(y+3)2=32(x+3)^2 + (y+3)^2 = 3^2

The distance between the centres is

AB=(1+3)2+(3+3)2=16+36=213AB = \sqrt{(1+3)^2 + (3+3)^2} = \sqrt{16+36} = 2\sqrt{13}

Since the circles touch externally,

r1+r2=ABr_1 + r_2 = AB

So,

3+r2=2133 + r_2 = 2\sqrt{13}

which gives

r2=213−3r_2 = 2\sqrt{13} - 3

The point of external contact divides the line joining the centres internally in the ratio of the radii. Therefore, using the section formula,

α=r1xB+r2xAr1+r2,β=r1yB+r2yAr1+r2\alpha = \frac{r_1 x_B + r_2 x_A}{r_1+r_2}, \qquad \beta = \frac{r_1 y_B + r_2 y_A}{r_1+r_2}

with A(−3,−3)A(-3,-3), B(1,3)B(1,3), r1=3r_1=3, and r2=213−3r_2=2\sqrt{13}-3.

So,

α=3(1)+(213−3)(−3)213,β=3(3)+(213−3)(−3)213\alpha = \frac{3(1) + (2\sqrt{13}-3)(-3)}{2\sqrt{13}}, \qquad \beta = \frac{3(3) + (2\sqrt{13}-3)(-3)}{2\sqrt{13}}

Simplifying,

α=12−613213,β=18−613213\alpha = \frac{12 - 6\sqrt{13}}{2\sqrt{13}}, \qquad \beta = \frac{18 - 6\sqrt{13}}{2\sqrt{13}}

Now,

β−α=(18−613)−(12−613)213=6213=313\beta - \alpha = \frac{(18 - 6\sqrt{13}) - (12 - 6\sqrt{13})}{2\sqrt{13}} = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}

Hence,

(β−α)2=(313)2=913(\beta - \alpha)^2 = \left(\frac{3}{\sqrt{13}}\right)^2 = \frac{9}{13}

Thus, m=9m=9 and n=13n=13.

Therefore, m+n=22m+n = 22. The correct option is C.

Using the line of centres

Given: The centre of C1C_1 is (−3,−3)(-3,-3) because a circle of radius 33 touching both axes in the third quadrant must be 33 units left of the yy-axis and 33 units below the xx-axis. The centre of C2C_2 is (1,3)(1,3).

Find: m+nm+n.

A common the solution's step incorrectly places the centre of C1C_1 at (3,3)(3,3). That is not consistent with the phrase third quadrant. The correct centre is (−3,−3)(-3,-3), and the correct working is based on that.

Since the circles touch externally, the contact point lies on the line segment joining the centres, and the radii to the point of contact are collinear. So the contact point divides ABAB in the ratio

AP:PB=r1:r2=3:(213−3)AP : PB = r_1 : r_2 = 3 : (2\sqrt{13}-3)

This directly leads to the section formula used above.

After finding

(β−α)2=913(\beta-\alpha)^2 = \frac{9}{13}

we identify m=9m=9 and n=13n=13, which are coprime. Therefore,

m+n=22m+n = 22

So the correct option is C.

Common mistakes

  • Placing the centre of C1C_1 at (3,3)(3,3) is wrong because the circle is stated to be in the third quadrant. A circle of radius 33 touching both axes there must have centre (−3,−3)(-3,-3). Always use the quadrant information before writing the centre.

  • Using the external-touch condition incorrectly can lead to impossible values such as a negative radius. For externally touching circles, the distance between centres equals the sum of radii, so here r2=AB−3r_2 = AB - 3, not any other expression.

  • Applying the section formula with the wrong endpoints or wrong ratio gives incorrect coordinates of P(α,β)P(\alpha,\beta). The point of contact lies on the line joining the centres and divides it internally in the ratio of the radii r1:r2r_1:r_2.

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