Two circles of different sizes are tangent at T. A is on the smaller circle, whereas B is on the larger one. Also, segment CD is tangent to the smaller circle, and crosses the goes through the larger circle and hits the other side at D. TD is a diameter, and AB || CD. If the radius of the smaller circle is 2, what is the radius of the larger one?

Question

Two circles of different sizes are tangent at T. A is on the smaller circle, whereas B is on the larger one. Also, segment CD is tangent to the smaller circle, and crosses the goes through the larger circle and hits the other side at D. TD is a diameter, and AB || CD. If the radius of the smaller circle is 2, what is the radius of the larger one?

I tried using perpendicular segments and right triangles, but they all seemed to fail. Could someone help? Thanks

Answer 1

To find the radius of the larger circle, we can use some geometric properties and relationships.

Let's start by labeling the points as shown in the diagram below:

```
C
|
|
T----D
|\ |
A | \
| \ \
| \ \
B
```

Given that TD is a diameter, we can conclude that angle TDC is a right angle because it is inscribed in a semicircle. Furthermore, since CD is tangent to the smaller circle, angle CTD is also a right angle.

Since AB is parallel to CD, angles ADC and ACD are congruent by alternate interior angles. Similarly, angles CTA and CTD are congruent by vertical angles.

Let r be the radius of the larger circle. Since TA is a radius of the smaller circle, its length is 2.

Now, let's use the information to set up an equation:

From right triangle CTD, we have:
CT^2 + TD^2 = CD^2

From right triangle CTA, we have:
CT^2 + TA^2 = AC^2

Substituting the known values and variables:
2^2 + (r + 2)^2 = (r + 4)^2

Expanding and simplifying the equation:
4 + r^2 + 4r + 4 = r^2 + 8r + 16

Combining like terms:
r^2 + 4r + 8 = r^2 + 8r + 16

Subtracting r^2 and 8r from both sides:
4 = 4r + 16

Subtracting 16 from both sides:
-12 = 4r

Dividing both sides by 4:
r = -3

Since a negative radius is not possible, we made an error somewhere.

Let's try a different approach:
Notice that angle CTA and angle ACD are vertical angles and therefore congruent. We can use this information to solve for the radius.

From right triangle CTA, we have:
CT^2 + TA^2 = CA^2

Substituting the known values and variables:
CT^2 + 2^2 = CA^2

From triangle ACD, we have:
AD = AC - CD

Substituting the known values and variables:
(r + 2) = CA - 2

From right triangle CTD, we have:
TD^2 + CD^2 = CT^2

Substituting the known values and variables:
(r + 4)^2 + r^2 = CT^2

Now, let's combine these equations to solve for r:

CT^2 + 2^2 = CA^2
(r + 4)^2 + r^2 = CT^2
(r + 2) = CA - 2

Expanding the equations:
(r + 4)^2 + r^2 = CT^2
r^2 + 8r + 16 + r^2 = CT^2
r + 2 = CA - 2

Combining like terms:
2r^2 + 8r + 16 = CT^2
4r = CA - 2

Substituting CA - 2 for r + 2:
2r^2 + 8r + 16 = (4r + 2)^2

Expanding the equation:
2r^2 + 8r + 16 = 16r^2 + 16r + 4

Subtracting 2r^2, 8r, and 4 from both sides:
0 = 14r^2 + 8r - 12

Dividing both sides by 2:
0 = 7r^2 + 4r - 6

Now, we can solve this quadratic equation to find the value of r. You can use the quadratic formula or factoring methods to solve this equation.

Using the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 7, b = 4, and c = -6.

Calculating the discriminant:
∆ = b^2 - 4ac
∆ = 4^2 - 4(7)(-6)
∆ = 16 + 168
∆ = 184

Applying the quadratic formula:
r = (-4 ± √184) / (2 × 7)
r ≈ (-4 ± 13.56) / 14

Evaluating both possible solutions:
r1 ≈ (-4 + 13.56) / 14
r1 ≈ 9.56 / 14
r1 ≈ 0.683

r2 ≈ (-4 - 13.56) / 14
r2 ≈ -17.56 / 14
r2 ≈ -1.25

Since the radius of a circle cannot be negative, we can conclude that the radius of the larger circle, in this case, is approximately 0.683.

Therefore, the radius of the larger circle is approximately 0.683.