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Magazine Chapter 7: Problem 47
Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$x^{2}-4 x+y^{2}+12 y=-4$$
Short Answer
Expert verified
Yes, it's a circle; center: (2, -6), radius: 6.
Step by step solution
01
Identify the Equation Components
The given equation is \(x^{2} - 4x + y^{2} + 12y = -4\). It resembles the standard form of a circle equation, which is \((x - h)^2 + (y - k)^2 = r^2\). We need to rewrite this equation to match the standard form, if possible, to determine if it represents a circle.
02
Complete the Square for x-terms
First, focus on the \(x\) terms: \(x^{2} - 4x\). We complete the square by adding and subtracting \(\left(\frac{-4}{2}\right)^2 = 4\). This transforms the expression into \((x-2)^2 - 4\).
03
Complete the Square for y-terms
Next, focus on the \(y\) terms: \(y^{2} + 12y\). We complete the square by adding and subtracting \(\left(\frac{12}{2}\right)^2 = 36\). This transforms the expression into \((y+6)^2 - 36\).
04
Rewrite the Equation
Substitute the completed squares back into the original equation: \((x-2)^2 - 4 + (y+6)^2 - 36 = -4\). Simplifying, we have:\((x - 2)^2 + (y + 6)^2 = 36\).
05
Determine the Center and Radius
The equation \((x - 2)^2 + (y + 6)^2 = 36\) is in the standard form of a circle's equation: \((x - h)^2 + (y - k)^2 = r^2\). The center \((h, k)\) is \((2, -6)\) and the radius \(r\) is \(\sqrt{36} = 6\).
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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 vital algebraic technique used to transform quadratic expressions into perfect square forms. This process makes it easier to interpret the equation and solve problems related to parabolas or circles.
- Start by focusing on quadratic terms, such as those seen in expressions like \(x^2 - 4x\) or \(y^2 + 12y\).
- To complete the square, identify the linear coefficient (the number in front of the linear term) and halve it. Then, square the result, which in these examples gives us \(\left(\frac{-4}{2}\right)^2 = 4\) and \(\left(\frac{12}{2}\right)^2 = 36\).
- Add and subtract this squared value within the equation to form a perfect square trinomial. For instance, \(x^2 - 4x\) becomes \((x-2)^2 - 4\), and \(y^2 + 12y\) turns into \((y+6)^2 - 36\).
Circle Equation
The general form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\).Here,
- \((h, k)\) represents the center of the circle, indicating its position on the coordinate plane.
- \(r\) is the radius, the constant distance from the center to any point on the circle's circumference.
Graphing Circles
Graphing a circle requires understanding its standard equation form and the resulting insights into its properties. The starting point is the circle equation \((x - h)^2 + (y - k)^2 = r^2\), from which we can easily extract important details.
- The center \((h, k)\) gives us the exact location around which the circle is plotted, effectively its middle point on a coordinate grid.
- The radius \(r\) indicates the circle's size, defining how large it spreads out from the center. Each point on the circle is exactly \(r\) units away from this center.
- Placing the center point on the graph.
- Using a compass or digital tool, define the constant radius distance, ensuring the circle remains equidistant from the center.
