Problem 29
Question
GEOMETRY A circle has a radius with endpoints at \((2,5)\) and \((-1,-4)\) . Find the circumference and area of the circle.
Step-by-Step Solution
Verified Answer
Radius = \(3\sqrt{10}\), circumference = \(6\pi\sqrt{10}\), area = \(90\pi\).
1Step 1: Calculate the Length of the Radius
Given the radius has endpoints at \((2,5)\) and \((-1,-4)\), use the distance formula to calculate the length of the radius: \(r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Substituting the values, \(r = \sqrt{(-1 - 2)^2 + (-4 - 5)^2} = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90}\), which can be simplified to \(3\sqrt{10}\).
2Step 2: Calculate the Circumference of the Circle
The circumference \(C\) of a circle is given by the formula \(C = 2\pi r\). Using \(r = 3\sqrt{10}\), the circumference is \(C = 2 \pi (3\sqrt{10}) = 6\pi \sqrt{10}\).
3Step 3: Calculate the Area of the Circle
The area \(A\) of a circle is calculated using the formula \(A = \pi r^2\). Substituting \(r = 3\sqrt{10}\), we find \(A = \pi (3\sqrt{10})^2 = \pi \times 90 = 90\pi\).
Key Concepts
Distance FormulaCircumference of a CircleArea of a Circle
Distance Formula
The distance formula helps in finding the distance between two points in a plane. It's like using a ruler on a graph to see how far apart the points are. When you have two points,
Imagine a right triangle formed by horizontal and vertical lines that connect the two points. The distance is the hypotenuse of that triangle. Applying it to our example, we have endpoints \( (2,5) \) and \( (-1,-4) \).
Substituting these into the formula, we calculate the radius of the circle as \( r = \sqrt{(-1 - 2)^2 + (-4 - 5)^2} = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90}\). This simplifies to \(3\sqrt{10}\), the length of the radius in simplest form, helping us in further calculations.
- Point 1: \(x_1, y_1\)
- Point 2: \(x_2, y_2\)
Imagine a right triangle formed by horizontal and vertical lines that connect the two points. The distance is the hypotenuse of that triangle. Applying it to our example, we have endpoints \( (2,5) \) and \( (-1,-4) \).
Substituting these into the formula, we calculate the radius of the circle as \( r = \sqrt{(-1 - 2)^2 + (-4 - 5)^2} = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90}\). This simplifies to \(3\sqrt{10}\), the length of the radius in simplest form, helping us in further calculations.
Circumference of a Circle
The circumference is the distance around a circle, similar to the perimeter for polygons. To find this, we use the formula \(C = 2\pi r\), where \(r\) is the radius and \(\pi\) is approximately 3.14159.
In our example, with a radius \(3\sqrt{10}\), the calculation becomes \(C = 2 \pi (3\sqrt{10})\). Expanding this, we have \(C = 6\pi \sqrt{10}\). It’s a simple multiplication that tells us how long a string would be if wrapped around the circle's edge.
Understanding circumference also provides insight into circular motion and various everyday scenarios involving wheels or circular tracks.
In our example, with a radius \(3\sqrt{10}\), the calculation becomes \(C = 2 \pi (3\sqrt{10})\). Expanding this, we have \(C = 6\pi \sqrt{10}\). It’s a simple multiplication that tells us how long a string would be if wrapped around the circle's edge.
Understanding circumference also provides insight into circular motion and various everyday scenarios involving wheels or circular tracks.
Area of a Circle
The area of a circle is a measure of the surface covered by the circle. The formula \(A = \pi r^2\) is used, where \(\pi\) (Pi) remains a constant, and \(r^2\) is the radius squared.
For our circle with radius \(3\sqrt{10}\), the calculation steps are:
Start thinking about the area as the amount of pizza topping on a round pizza, stretching from the center to the edge.
For our circle with radius \(3\sqrt{10}\), the calculation steps are:
- First, square the radius: \( (3\sqrt{10})^2 = 9 \times 10 = 90\)
- Then multiply by \(\pi\): \(A = \pi \times 90 = 90\pi\)
Start thinking about the area as the amount of pizza topping on a round pizza, stretching from the center to the edge.
Other exercises in this chapter
Problem 29
Find the center and radius of the circle with the given equation. Then graph the circle. $$ x^{2}+y^{2}-18 x-18 y+53=0 $$
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Write an equation for the ellipse that satisfies each set of conditions. Write the equation \(10 x^{2}+2 y^{2}=40\) in standard form.
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What are the coordinates of the points that lie on the graphs of both \(x^{2}+y^{2}=25\) and \(2 x^{2}+3 y^{2}=66 ?\)
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Find an equation for a hyperbola centered at the origin with a horizontal transverse axis of length 8 units and a conjugate axis of length 6 units.
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