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Chapter 10: Problem 17

Write an equation for each circle described below. a circle with center at \((-3,6)\) and a radius with endpoint at \((0,6)\)

Short Answer

Expert verified
The equation of the circle is \((x + 3)^2 + (y - 6)^2 = 9\).

Step by step solution

01

Find the Radius

To find the radius, calculate the distance between the center of the circle, \((-3, 6)\), and the endpoint on the circle, \((0, 6)\). Since these points differ only in their x-coordinates, the distance is the difference in x-coordinates. Thus, the radius, \(r\), is \(|0 - (-3)| = 3\).
02

Identify Circle Equation Form

The general equation for a circle with a center at \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). In this case, the center is \((-3, 6)\) and the radius is \(3\).
03

Substitute Values into Circle Equation

Substitute the center coordinates and radius into the circle equation form. With \(h = -3\), \(k = 6\), and \(r = 3\), the equation becomes:\[(x + 3)^2 + (y - 6)^2 = 3^2\].
04

Simplify the Equation

Calculate \(r^2\), which is \(3^2 = 9\). So the equation simplifies to:\[(x + 3)^2 + (y - 6)^2 = 9\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radius Calculation
The radius of a circle is crucial for defining the circle's dimension. In our exercise, the circle has a center at \((-3, 6)\) and an endpoint on the circle at \((0, 6)\). The radius is simply the distance between the center and the endpoint. Since the y-coordinates of these points are identical, we only consider the x-coordinates. To find this distance, subtract the x-coordinate of the center from the x-coordinate of the endpoint.
  • The distance formula is essentially: \(|x_2 - x_1|\).
  • In our problem, it's calculated as: \(|0 - (-3)| = 3\).
Thus, the radius of the given circle is \(3\). Calculating the radius accurately is vital because it will be used in the circle's equation to reflect its true size.
Circle Center
A circle's center is represented by the point \((h, k)\) in its equation. This point is fundamental because it establishes where the circle rests in a coordinate plane. In our exercise, the center is given as \((-3, 6)\).
  • The x-coordinate of the center is \(-3\).
  • The y-coordinate of the center is \(6\).
The center not only helps in determining the circle's placement but also plays a critical role in its equation. By knowing the center, you can solve multiple problems related to positioning and symmetry of the circle.
Geometric Equations
The equation of a circle is a classic form in geometry. It is derived from the Pythagorean Theorem and is written as:\[(x - h)^2 + (y - k)^2 = r^2\]This equation connects the circle's geometric properties: the center \((h, k)\) and radius \(r\).
  • For our circle, the center \((h, k)\) is \((-3, 6)\).
  • The radius \(r\) is calculated as \(3\).
To form the equation for this circle, substitute the center coordinates and radius value:\[(x + 3)^2 + (y - 6)^2 = 9\]The equation reveals all necessary details about the circle's size and its position on the coordinate plane. It is a foundational concept in geometry and crucial for both theoretical and applied mathematics.

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