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Magazine Chapter 10: Problem 36
Write each equation of a circle in standard form and graph it. Give the coordinates of its center and give the radius. $$ x^{2}+y^{2}+6 x-4 y=-12 $$
Short Answer
Expert verified
The center is (-3, 2) and the radius is 1.
Step by step solution
01
Identify the Standard Form of a Circle
The standard form of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius.
02
Rearrange the Equation
Start with the given equation: \[ x^2 + y^2 + 6x - 4y = -12 \] Move the constant term to the right side: \[ x^2 + y^2 + 6x - 4y + 12 = 0 \] Rearrange it: \[ x^2 + 6x + y^2 - 4y = -12 \]
03
Complete the Square for the x-terms
Identify the terms for completing the square: \[ x^2 + 6x \] Take half of the coefficient of \(x\) (which is 6), square it, and add to both sides: \[ (6/2)^2 = 9 \] Add 9 to both sides of the equation: \[ x^2 + 6x + 9 \] This becomes \((x + 3)^2\).
04
Complete the Square for the y-terms
Identify the terms for completing the square: \[ y^2 - 4y \] Take half of the coefficient of \(y\) (which is -4), square it, and add to both sides: \[ (-4/2)^2 = 4 \] Add 4 to both sides of the equation: \[ y^2 - 4y + 4 \] This becomes \((y - 2)^2\).
05
Write the Equation in Standard Form
After completing the square for both \(x\) and \(y\), the equation becomes: \[ (x + 3)^2 + (y - 2)^2 = 1 \] This is the equation of the circle in standard form.
06
Find the Center and Radius
From the standard form equation: \[ (x + 3)^2 + (y - 2)^2 = 1 \] The center of the circle \((h, k)\) is \((-3, 2)\). The radius \(r\) is the square root of 1, so \(r = 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form of a Circle
The standard form of a circle is a way to express the equation of a circle so that its center and radius are easily identifiable. This form is written as \((x-h)^2 + (y-k)^2 = r^2\). In this equation:
- \(h\) and \(k\) are the coordinates of the center of the circle.
- \(r\) is the radius, which is the distance from the center to any point on the circle.
Completing the Square
Completing the square is a mathematical technique used to convert a quadratic equation into a format that reveals its key features. For circles, it鈥檚 used to transform separate \(x\) and \(y\) terms into perfect square trinomials. Here's how that works:
- Identify the quadratic and linear terms for \(x\) and \(y\). For example, \(x^2 + 6x\) needs to be completed to become a perfect square trinomial.
- Take half of the linear term's coefficient, square it, and add it inside the equation. For \(x^2 + 6x\), half of 6 is 3, and squaring it gives 9, transforming it into \((x+3)^2\).
- Balance the equation by adding the same value to the opposite side if modifying a whole equation, maintaining equality.
Center of a Circle
The center of a circle is one of its most fundamental parts, located at the coordinate point \((h, k)\) in the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\). By transforming the general equation into this form:
- The \(h\) value is found in \((x-h)^2\), meaning it is essential to keep the sign in mind as \(h\) might appear as - in the transformed equation. For example, \((x+3)^2\) implies \(h = -3\).
- The \(k\) value is found in \((y-k)^2\), similarly. If the equation reads \((y-2)^2\), \(k\) is actually 2, as it maintains a negative sign when isolated from the square.
Radius of a Circle
The radius of a circle is the distance from its center to any point on its circumference. In the standard form of the circle equation \((x-h)^2 + (y-k)^2 = r^2\), the radius is extracted from the \(r^2\) term:
- The value \(r^2\) is on the right-hand side of the equality, which is the square of the radius.
- To find the actual radius \(r\), you take the square root of \(r^2\). For instance, if the equation is \((x+3)^2 + (y-2)^2 = 1\), then \(r = \sqrt{1} = 1\).
