Question

Two circles have circumferences that add up to \(12 \pi\) centimeters and areas that add up to \(20 \pi\) square centimeters. Find the radius of each circle.

Short answer

The radii of the circles are 2 cm and 4 cm.

Step-by-step solution

- Understand the problem

Identify that you need to find the radii of two circles given their combined circumferences and areas.

- Write down the formulas

Recall the formulas for circumference and area of a circle. Circumference: \(C = 2 \pi r\). Area: \(A = \pi r^2\).

- Set up equations

Let the radii of the circles be \(r_1\) and \(r_2\). Therefore, their circumferences are \(2 \pi r_1\) and \(2 \pi r_2\). Since their combined circumference is \(12 \pi\): \(2 \pi r_1 + 2 \pi r_2 = 12 \pi\). Their areas are \(\pi r_1^2\) and \(\pi r_2^2\). Since their combined area is \(20\pi\): \(\pi r_1^2 + \pi r_2^2 = 20 \pi\).

- Simplify the equations

Divide both sides of the first equation by \(2 \pi\) and the second equation by \(\pi\): \(r_1 + r_2 = 6\), \(r_1^2 + r_2^2 = 20\).

- Use substitution

Express one radius in terms of the other using the first equation: \(r_2 = 6 - r_1\). Substitute \(r_2\) in the second equation: \(r_1^2 + (6 - r_1)^2 = 20\).

- Expand and simplify

Expand and simplify the equation: \(r_1^2 + (36 - 12r_1 + r_1^2) = 20\), \(2r_1^2 - 12r_1 + 36 = 20\), \(2r_1^2 - 12r_1 + 16 = 0\).

- Solve the quadratic equation

Solve the quadratic equation \(r_1^2 - 6r_1 + 8 = 0\) using the quadratic formula \(r = \frac{-b \,\pm \, \sqrt{b^2 - 4ac}}{2a}\): \(r_1 = \frac{6 \,\pm \, \sqrt{36 - 32}}{2}\), \(r_1 = \frac{6 \,\pm \, 2}{2}\). So, \(r_1 = 4\) or \(r_1 = 2\).

- Determine the radii

Using \(r_1 = 4\), we find \(r_2 = 6 - 4 = 2\). Using \(r_1 = 2\), we find \(r_2 = 6 - 2 = 4\).

Key concepts

Circumference of a Circle

The circumference of a circle is the distance around the circle's edge, which can be likened to the perimeter of a circle. To find the circumference, we use the formula: \(C = 2 \pi r\). Here, \(r\) represents the radius of the circle, and \(\pi\) (pi) is a mathematical constant approximately equal to 3.14159.

For example, if a circle has a radius of 3 cm, its circumference would be calculated as: \[C = 2 \pi \times 3 = 6 \pi \text{cm}\]. This makes it easy to compute the distance around any circle if the radius is known.

In the exercise, two circles have a combined circumference of \12 \pi\. By setting up the equation \2 \pi r_1 + 2 \pi r_2 = 12 \pi\, we can solve for the sum of the radii.

Area of a Circle

The area of a circle represents the space enclosed within its circumference. It can be found using the formula: \A = \pi r^2\. Here, \r\ is the radius, and as before, \pi\ is approximately 3.14159.

Imagine a circle with a radius of 3 cm. To find its area, we calculate: \[A = \pi \times (3)^2 = 9 \pi \text{square cm}\]. This formula gives us the precise measurement of the space within the circle.

In this exercise, the areas of two circles add up to \20 \pi \text{square cm}\. By setting up the equation \ \pi r_1^2 + \pi r_2^2 = 20 \pi\, we can solve for the areas involving the radii.

Quadratic Equations

A quadratic equation is a second-degree polynomial equation in a single variable \x\, with the form \[ax^2 + bx + c = 0\]. Solutions to the quadratic equation can be found using the quadratic formula: \[x = \frac{-b \pm \, \sqrt{b^2 - 4ac}}{2a}\].

In the given exercise, we derived a quadratic equation \2r_1^2 - 12r_1 + 16 = 0\ and then simplified it to \[r_1^2 - 6r_1 + 8 = 0\]. To solve for \r_1\, we used the quadratic formula: \r_1 = \frac{6 \pm \, \sqrt{36 - 32}}{2}= 4\ or \2\. By solving the quadratic equation, we found possible values for the radius of the first circle.

Radius of a Circle

The radius of a circle is the distance from the center point to any point on the edge of the circle. It is a key component used in almost all circle-related formulas, such as the circumference and area formulas.

For instance, in the exercise, if \r_1\ represents the radius of the first circle and \r_2\ represents the radius of the second circle, we used the relationship \[r_1 + r_2 = 6\]. By substituting one radius in terms of the other: \[r_2 = 6 - r_1\], we were able to simplify the equations further.

Finally, we determined the possible radii to be \[r_1 = 4\ and \r_2 = 2\, or \r_1 = 2\ and \r_2 = 4\]. Understanding the radius is essential, as it connects the physical dimensions of the circle to its geometrical properties.