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Magazine Chapter 10: Problem 29
Find the center and radius of the circle with the given equation. Then graph the circle. $$ x^{2}+y^{2}-18 x-18 y+53=0 $$
Short Answer
Expert verified
The circle's center is (9, 9) and the radius is \(\sqrt{109}\).
Step by step solution
01
Rearrange the Equation
First, rearrange the given equation by grouping the terms with the same variables together: \\[x^2 - 18x + y^2 - 18y + 53 = 0\] \Next, remember that the general form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
02
Complete the Square for x-terms
To complete the square for the \(x\)-terms, take the coefficient of \(x\), divide it by 2, and square it. \Coefficient of \(x\) = -18 \\((-18 / 2)^2 = 81\) \Add and subtract 81 inside the equation to complete the square: \\[x^2 - 18x + 81\] \This can be rewritten as \((x - 9)^2\).
03
Complete the Square for y-terms
Similarly, complete the square for the \(y\)-terms: \Coefficient of \(y\) = -18 \\((-18 / 2)^2 = 81\) \Add and subtract 81 inside the equation to complete the square: \\[y^2 - 18y + 81\] \This can be rewritten as \((y - 9)^2\).
04
Simplify the Equation and Find the Center and Radius
Substitute the completed squares back into the equation and simplify: \\[ (x - 9)^2 + (y - 9)^2 - 81 - 81 + 53 = 0\] \\[ (x - 9)^2 + (y - 9)^2 - 109 = 0\] \\[ (x - 9)^2 + (y - 9)^2 = 109\] \The equation is now in the standard form \((x-h)^2 + (y-k)^2 = r^2\). \Thus, the center \((h, k)\) of the circle is \((9, 9)\) and the radius \(r\) is \(\sqrt{109}\).
05
Graph the Circle
To graph the circle, plot the center at point \((9, 9)\) on the coordinate plane. Then, use the radius \(\sqrt{109}\), which is approximately 10.44, to draw the circle. Start from the center and mark points approximately 10.44 units away in all directions to sketch the circle accurately.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This technique is extremely useful in converting equations into a form that is easier to work with for certain types of problems. For a quadratic expression in the form \( ax^2 + bx + c \), you start by focusing on the \( x \) terms.
- Take the coefficient of \( x \), divide it by 2, and then square the result.
- This result is then added and subtracted within the expression to maintain its balance.
- This transforms \( x^2 - 18x \) into \((x - 9)^2 - 81\).
Standard Form of a Circle
The standard form of a circle's equation helps to identify the circle's key characteristics easily. A circle's equation in standard form is written as \((x - h)^2 + (y - k)^2 = r^2\). Here:
- \((h, k)\) represents the circle's center.
- \(r\) stands for the radius of the circle.
- The center is at \((9, 9)\).
- The radius is \(\sqrt{109}\). This transformation simplifies graphing and understanding the circle's properties.
Graphing Circles
Graphing a circle starts with identifying its center and radius from its equation. Once the equation is in standard form \((x - h)^2 + (y - k)^2 = r^2\), the process becomes straightforward. Follow these easy steps:
- Plot the center \((h, k)\) on the coordinate plane. This serves as the pivot point.
- From the center, mark points that are \(r\) units away in every direction.
Center and Radius of a Circle
Understanding the center and radius of a circle is crucial for graphing and analyzing its properties. Given a circle's equation in standard form \((x - h)^2 + (y - k)^2 = r^2\):
- The center point \((h, k)\) serves as the circle's anchor. It dictates where the circle is located on the graph.
- The radius \(r\) measures the distance from the center to any point on the circle's edge. It defines the circle's size.
