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Ch. 2 - Graphs and Functions
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 3, Problem 29

Give the center and radius of the circle represented by each equation. x2+y2-4x+12y=-4

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1
Start with the given equation of the circle: \(x^2 + y^2 - 4x + 12y = -4\).
Group the \(x\) terms and \(y\) terms together: \((x^2 - 4x) + (y^2 + 12y) = -4\).
Complete the square for the \(x\) terms: take half of the coefficient of \(x\) (which is \(-4\)), divide by 2 to get \(-2\), then square it to get \(4\). Add and subtract \(4\) inside the equation.
Complete the square for the \(y\) terms: take half of the coefficient of \(y\) (which is \(12\)), divide by 2 to get \(6\), then square it to get \(36\). Add and subtract \(36\) inside the equation.
Rewrite the equation as perfect square trinomials: \((x - 2)^2 - 4 + (y + 6)^2 - 36 = -4\). Then, move the constants to the right side and simplify to get the standard form of the circle equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Standard Form of a Circle Equation

The standard 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. Converting a general quadratic equation into this form helps identify these key features directly.
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Circles in Standard Form

Completing the Square

Completing the square is a method used to rewrite quadratic expressions as perfect square trinomials. This technique is essential for transforming the given equation into the standard form of a circle by grouping and completing squares for x and y terms.
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Identifying the Center and Radius from the Equation

Once the equation is in standard form, the center (h, k) is found by taking the opposite sign of the values inside the parentheses, and the radius r is the square root of the constant on the right side. This allows direct extraction of the circle's geometric properties.
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Circles in Standard Form