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Chapter 13: Problem 30

Find an equation of the circle satisfying the given conditions. Center \((0,0),\) radius 5

Short Answer

Expert verified
The equation is \[ x^2 + y^2 = 25 \].

Step by step solution

01

Understand the General Equation of a Circle

The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by \[ (x - h)^2 + (y - k)^2 = r^2 \].
02

Plug in the Center Coordinates

Given the center is at \( (0,0) \), substitute \ h = 0 \ and \ k = 0 \ into the equation: \[ (x - 0)^2 + (y - 0)^2 = r^2 \], which simplifies to \[ x^2 + y^2 = r^2 \].
03

Substitute the Radius

The radius is given as 5. Substitute \ r = 5 \ into the equation \[ x^2 + y^2 = 5^2 \].
04

Simplify the Equation

Simplify \[ 5^2 \] to get 25. Therefore, the equation becomes \[ x^2 + y^2 = 25 \].

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Key Concepts

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

radius
The radius of a circle is the distance from its center to any point on the circumference. In this exercise, we were given that the radius is 5. Understanding the radius is crucial because it helps determine the size of the circle. Think of the radius as a constant measurement that stretches out from the center in every direction. If you were to draw a circle using a compass, the distance between the compass point (the center) and the pencil tip (on the circle's edge) is the radius. In our problem, the radius was 5, which is then squared to become a part of the equation.
center coordinates
The center coordinates of a circle pinpoint the exact middle point of the circle. Given coordinates \(h, k\), where \(h\) is the x-coordinate and \(k\) is the y-coordinate, they act as the anchor for the circle. For the exercise mentioned, the center is at \(0, 0\). This simplifies our equation significantly because there is no need to account for any shift in the circle away from the origin. So, when the center is at the origin, our generic equation \((x - h)^2 + (y - k)^2 = r^2\) simplifies greatly to \(x^2 + y^2 = r^2\), as \(h = 0 \) and \( k = 0 \).
general equation of a circle
The general equation of a circle encapsulates the key information about a circle: its center and radius. This equation is \((x - h)^2 + (y - k)^2 = r^2\), where \(h\) and \(k\) are the center coordinates, and \(r\) is the radius. This formula is derived from the distance formula. It ensures that every point \(x, y\) on the circle is equidistant from the center. Plugging the specific center and radius into this equation customizes it for any circle. Here, with a center at \(0, 0\) and radius 5, the equation simplifies beautifully to \(x^2 + y^2 = 25\).
simplifying equations
Simplifying equations involves reducing them to their most basic form. In our circle equation, each step simplified it. Initially, we started with \((x - h)^2 + (y - k)^2 = r^2\). Substituting \h = 0\ and \k = 0\ simplified it to \(x^2 + y^2 = r^2\). Next, using the given radius of 5, we squared the radius, turning \(r^2\) into 25. This final step gave us the simplified equation \x^2 + y^2 = 25 \. Simplifying makes equations easier to work with and interpret, reducing potential computational errors and making visualization straightforward.

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Most popular questions from this chapter

Find an equation of a circle satisfying the given conditions. The endpoints of a diameter are \((7,3)\) and \((-1,-3)\)

Why should a graphing calculator's window be "squared" before graphing a circle?

Solve using a graphing calculator. Round all values to two decimal places. $$ \begin{aligned} &4 x y-7=0\\\ &x-3 y-2=0 \end{aligned} $$

Match the equation with the center or vertex of its graph, listed in the column on the right. a) Vertex: \((-2,5)\) b) Vertex: \((5,-2)\) c) Vertex: \((2,-5)\) d) Vertex: \((-5,2)\) e) Center: \((-2,5)\) f) Center: \((2,-5)\) g) Center: \((5,-2)\) h) Center: \((-5,2)\) $$(x+5)^{2}+(y-2)^{2}=9$$

Solve $$ \begin{aligned} &p^{2}+q^{2}=13,\\\ &\frac{1}{p q}=-\frac{1}{6} \end{aligned} $$
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