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Chapter 1: Problem 56

Determine the center and radius of the circle with the given equation. $$x^{2}+(y-12)^{2}=1$$

Short Answer

Expert verified
The center of the circle with given equation is \((0, 12)\) and the radius is \(1\)

Step by step solution

01

Identify the Center

The general form of a circle equation is \(x^{2}+(y-h)^{2}=r^{2}\), where \((h, k)\) are the coordinates of the center. By comparing with the given equation \(x^{2}+(y-12)^{2}=1\), it can be seen that it's in similar form, meaning that the coordinates of center of the circle are \((h, k) = (0, 12)\)
02

Identify the Radius

The radius \(r\) of the circle in the general equation is given by the square root of the constant in the equation. Here, instead of \(r^{2}\) in the general equation, there is '1' in the given equation. So the radius \(r\) of the circle is \(\sqrt{1} = 1\)

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Key Concepts

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

Center of a Circle
Understanding the center of a circle is crucial in the study of geometry. In the equation of a circle, the center is represented as \(h, k\). The center refers to the exact middle point of the circle, which is equidistant from any point on the circle's edge. Knowing the center helps in accurately positioning the circle in a coordinate plane.To find the center in a circle equation, \(x^2 + (y - h)^2 = r^2\), we locate the values of \(h\) and \(k\). Here鈥檚 how:
  • Identify the parts of the equation that correspond to \(h\) and \(k\).
  • The term \(x - h\) indicates the horizontal shift from the origin, giving the x-coordinate.
  • The term \(y - k\) shows the vertical shift, providing the y-coordinate.
In the given problem, \(x^2 + (y - 12)^2 = 1\), we deduced that the equation can be rearranged to highlight \((0, 12)\) as the center. Understanding this helps map the circle correctly on the graph.
Radius of a Circle
The radius of a circle is a fundamental concept that defines the distance from the center to any point on the circle's edge. It is a fixed length that remains constant for a given circle, making it essential for calculating the circle's circumference and area.In a standard circle equation, \(x^2 + (y-h)^2 = r^2\), the radius \(r\) is found as follows:
  • Compare the equation with the general form to find the term that equals \('r^2'\).
  • Take the square root of this term to solve for the radius, \(r\).
For the equation \(x^2 + (y-12)^2 = 1\), the number \('1'\) is equivalent to \(r^2\). Therefore, the radius \(r\) is \(\sqrt{1} = 1\). Knowing the radius helps in creating accurate circle representations and performing geometric calculations.
General Form of a Circle Equation
The general form of a circle equation is a standard format used to represent circles mathematically. It typically looks like this: \(x^2 + (y-h)^2 = r^2\). This form is universal, making it easy to apply to different scenarios involving circles.Here are the essential components of this form:
  • \(x\) and \(y\) denote the variables that plot the circle on a coordinate plane.
  • \(h\) and \(k\) specify the coordinates for the center of the circle, providing its precise position.
  • \(r^2\) represents the radius squared, giving insight into the size of the circle.
By converting an equation into this form, you can easily identify both the center and radius of the circle, as demonstrated in \(x^2 + (y-12)^2 = 1\). The conversion of a circle equation into its general form is vital for analyzing and solving circle-related problems efficiently, providing clarity and an orderly approach to geometry.

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Most popular questions from this chapter

The notation \(\left.f(x)\right|_{a} ^{b}\) is used to denote the difference \(f(b)-f(a) .\) That is, $$\left.f(x)\right|_{a} ^{b}=f(b)-f(a)$$ Evaluate \(\left.f(x)\right|_{0} ^{b}\) for the given function \(f\) and the indicated values of \(a\) and \(b\). $$f(x)=\sqrt{8-x} ;\left.f(x)\right|_{0} ^{8}$$

Evaluate \(\sqrt{a^{2}+b^{2}}\) when \(a=-3\) and \(b=4 .[\text { A. } 1]\)

Find the product of a nonzero number and its negative reciprocal.

solve by completing the square or by using the quadratic formula. $$3 x^{2}-5 x-4=0$$

Use interval notation to express the solution set of each inequality. $$|x+4|<2$$
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