Maths Q12 – JEE-MAIN 2025

Let

C_{1}

be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let

C_{2}

be the circle with centre

(1, 3)

that touches

C_{1}

externally at the point

(α, β)

. If

(β - α)^{2} = \frac{m}{n}

gcd (m, n) = 1

, then

m + n

is equal to

Clue 1 of 3: Circle Properties

Determine the center and radius of $C_{1}$ from its description, noting it's in the third quadrant and touches both axes.

🧠 Full Solution Path

Step 1: Determine properties of

C_{1}

and

C_{2}

and their radii.✦ Active

Circle $C_{1}$ is in the third quadrant, has radius $r_{1} = 3$ , and touches both coordinate axes. Therefore, its center is $O_{1} = (- 3, - 3)$ . Circle $C_{2}$ has center $O_{2} = (1, 3)$ . Let its radius be $r_{2}$ . Since $C_{1}$ and $C_{2}$ touch externally, the distance between their centers $O_{1} O_{2}$ is equal to the sum of their radii $r_{1} + r_{2}$ .

O_{1} O_{2} = \sqrt{(1 - (- 3))^{2} + (3 - (- 3))^{2}} = \sqrt{(4)^{2} + (6)^{2}} = \sqrt{16 + 36} = \sqrt{52}

Thus, $r_{1} + r_{2} = \sqrt{52} ⟹ 3 + r_{2} = \sqrt{52} ⟹ r_{2} = \sqrt{52} - 3$ .

Step 2: Find the coordinates of the point of tangency

(α, β)

.○ Expand

The point of tangency $(α, β)$ divides the line segment $O_{1} O_{2}$ in the ratio $r_{1} : r_{2} = 3 : (\sqrt{52} - 3)$ internally. Using the section formula:

α = \frac{r_{2} x_{1} + r_{1} x_{2}}{r_{1} + r_{2}} = \frac{(\sqrt{52} - 3) (- 3) + 3 (1)}{\sqrt{52}} = \frac{- 3 \sqrt{52} + 9 + 3}{\sqrt{52}} = \frac{12 - 3 \sqrt{52}}{\sqrt{52}}

β = \frac{r_{2} y_{1} + r_{1} y_{2}}{r_{1} + r_{2}} = \frac{(\sqrt{52} - 3) (- 3) + 3 (3)}{\sqrt{52}} = \frac{- 3 \sqrt{52} + 9 + 9}{\sqrt{52}} = \frac{18 - 3 \sqrt{52}}{\sqrt{52}}

💡 Teacher's Secret Hint

Remember to use the correct ratio for internal division.

Step 3: Calculate

(β - α)^{2}

and

m + n

.○ Expand

Now, calculate the difference $β - α$ :

β - α = (\frac{18 - 3 \sqrt{52}}{\sqrt{52}}) - (\frac{12 - 3 \sqrt{52}}{\sqrt{52}}) = \frac{18 - 3 \sqrt{52} - 12 + 3 \sqrt{52}}{\sqrt{52}} = \frac{6}{\sqrt{52}}

Next, calculate $(β - α)^{2}$ :

(β - α)^{2} = {(\frac{6}{\sqrt{52}})}^{2} = \frac{36}{52}

Simplify the fraction: $\frac{36}{52} = \frac{9 \times 4}{13 \times 4} = \frac{9}{13}$ . Given $(β - α)^{2} = \frac{m}{n}$ and $gcd (m, n) = 1$ , we have $m = 9$ and $n = 13$ . Therefore, $m + n = 9 + 13 = 22$ .

💡 Teacher's Secret Hint

Ensure the fraction is simplified to its lowest terms to correctly identify $m$ and $n$ where $gcd (m, n) = 1$ .

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Maths Question 12 – JEE-MAIN 2025

Clue 1 of 3: Circle Properties

🧠 Full Solution Path