Question

Two circles are drawn, with the same center. A chord of the large circle is drawn, so that it is tangent to the small circle. If the chord has a length of 10, then find the area of the ring-shaped region that is inside the large circle but outside the small circle.

Answer 1

To solve the problem, we need to find the areas of the two concentric circles and then compute the area of the ring-shaped region between them.

Let \( r \) be the radius of the small circle and \( R \) be the radius of the large circle. The given chord of the large circle is tangent to the small circle and has a length of 10.

### Step 1: Relationship between the chord and the circles

Since the chord is tangent to the small circle, it will be at a distance equal to the radius of the small circle \( r \) from the center of the circles. Let the perpendicular distance from the center of the circle to the chord be \( r \).

Let \( d \) be the distance from the center to the chord along the radius of the large circle. By the properties of chords in a circle, we know that:

\[
d = \sqrt{R^2 - \left( \frac{L}{2} \right)^2}
\]

where \( L \) is the length of the chord. We know the length \( L = 10 \), so:

\[
\frac{L}{2} = 5
\]

Substituting into the equation gives:

\[
d = \sqrt{R^2 - 5^2} = \sqrt{R^2 - 25}
\]

Since this distance \( d \) must also equal the radius \( r \) of the small circle, we can set up the equation:

\[
r = \sqrt{R^2 - 25}
\]

### Step 2: Area calculations

The area of the large circle \( A_R \) is given by:

\[
A_R = \pi R^2
\]

The area of the small circle \( A_r \) is given by:

\[
A_r = \pi r^2
\]

To find the area of the ring-shaped region, we compute:

\[
\text{Area of the ring} = A_R - A_r = \pi R^2 - \pi r^2 = \pi (R^2 - r^2)
\]

### Step 3: Determining \( R^2 - r^2 \)

We know that:

\[
r^2 = R^2 - 25
\]

Thus,

\[
R^2 - r^2 = R^2 - (R^2 - 25) = 25
\]

### Step 4: Computing the area of the ring

Substituting \( R^2 - r^2 = 25 \) into the area of the ring, we have:

\[
\text{Area of the ring} = \pi (R^2 - r^2) = \pi (25) = 25\pi
\]

### Final answer:

Hence, the area of the ring-shaped region that is inside the large circle but outside the small circle is

\[
\boxed{25\pi}
\]