CAT 2024 Slot 2 QA Question 13

Okay, so I'm trying to solve this CAT 2024 question about circles touching each other. We have three circles of equal radii that touch each other externally. Then, two more circles, X and Y, are drawn such that both touch each of the three previous circles. The radius of X is more than Y, and we need to find the ratio of their radii.

Hmm, okay. So, three equal circles touching each other externally. I imagine them forming a sort of triangle, each touching the other two. Then, circles X and Y are each touching all three of these circles. Since X has a larger radius than Y, I think X must be the outer circle and Y the inner one. But wait, actually, in such configurations, there are two possible circles that can touch all three: one inside the triangle formed by the three circles (the inner Soddy circle) and one outside (the outer Soddy circle). So, X is the outer one and Y is the inner one.

I remember something about Soddy circles and Descartes' Circle Theorem. Maybe that's the key here. Let me recall Descartes' Theorem. It states that if four circles are mutually tangent, their curvatures satisfy a certain equation.

Curvature is defined as the reciprocal of the radius, right? So, if k = 1/r. For four circles with curvatures k1, k2, k3, k4, the theorem says:

k4 = k1 + k2 + k3 ± 2√(k1k2 + k2k3 + k3k1)

But in our case, we have three equal circles. Let's denote their radii as r. So, their curvatures are all k = 1/r.

We need to find the curvatures of the fourth circle, which can be either the inner or outer Soddy circle. Since X is larger, its curvature will be smaller (because curvature is inversely proportional to radius). Y will have a larger curvature.

Wait, actually, the outer Soddy circle will have a smaller curvature (since it's larger) and the inner one will have a larger curvature. So, if X is the outer circle, its curvature will be k4 = k1 + k2 + k3 - 2√(k1k2 + k2k3 + k3k1). And Y will be the inner one, so its curvature will be k4 = k1 + k2 + k3 + 2√(k1k2 + k2k3 + k3k1).

But let me make sure. Descartes' formula gives two solutions for the fourth circle: one adding and one subtracting the square root term. The one with the plus sign gives the inner circle, which has a larger curvature (smaller radius), and the one with the minus sign gives the outer circle, which has a smaller curvature (larger radius). So, in our case, X is the outer circle, so we'll use the minus sign, and Y is the inner circle, using the plus sign.

So, Computing this step by step.

First, let's denote the radius of the three equal circles as r. So, their curvature is k = 1/r.

Now, applying Descartes' Theorem for the outer circle X:

k_X = k1 + k2 + k3 - 2√(k1k2 + k2k3 + k3k1)

Since all k1, k2, k3 are equal to 1/r, let's substitute:

k_X = (1/r) + (1/r) + (1/r) - 2√[(1/r)(1/r) + (1/r)(1/r) + (1/r)(1/r)]

Simplify the terms:

k_X = 3/r - 2√[3*(1/r^2)]

Simplify the square root:

√[3*(1/r^2)] = √3 / r

So, substituting back:

k_X = 3/r - 2*(√3 / r) = (3 - 2√3)/r

Therefore, the curvature of X is (3 - 2√3)/r, so the radius of X is r_X = r / (3 - 2√3)

Similarly, for the inner circle Y, we use the plus sign:

k_Y = 3/r + 2√3 / r = (3 + 2√3)/r

Thus, the radius of Y is r_Y = r / (3 + 2√3)

Now, we need the ratio r_X : r_Y.

So, Computing r_X / r_Y:

(r / (3 - 2√3)) / (r / (3 + 2√3)) = (3 + 2√3) / (3 - 2√3)

To simplify this ratio, we can rationalize the denominator.

Multiply numerator and denominator by (3 + 2√3):

[(3 + 2√3)^2] / [(3)^2 - (2√3)^2]

Compute numerator:

(3 + 2√3)^2 = 3^2 + 232√3 + (2√3)^2 = 9 + 12√3 + 12 = 21 + 12√3

Denominator:

9 - (4*3) = 9 - 12 = -3

So, the ratio becomes (21 + 12√3)/(-3) = -(21 + 12√3)/3 = -7 -4√3

But since ratio is positive, we take absolute value: 7 + 4√3

Therefore, the ratio r_X : r_Y is (7 + 4√3) : 1

Looking at the options, that's option B.

First, applying Descartes' Theorem correctly: yes, for four mutually tangent circles, the formula is correct. Since all three initial circles have equal radii, their curvatures are equal, so the formula simplifies as I did.

Calculating k_X: 3/r - 2√3/r = (3 - 2√3)/r, correct.

Similarly, k_Y: 3/r + 2√3/r = (3 + 2√3)/r, correct.

Then, r_X = r / (3 - 2√3), r_Y = r / (3 + 2√3). So, the ratio is (3 + 2√3)/(3 - 2√3). Rationalizing:

Multiply numerator and denominator by (3 + 2√3):

Numerator: (3 + 2√3)^2 = 9 + 12√3 + 12 = 21 + 12√3

Denominator: (3)^2 - (2√3)^2 = 9 - 12 = -3

So