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Magazine Chapter 0: Problem 54
Find the center and radius of each circle. $$x^{2}+y^{2}-8 x-6 y+21=0$$
Short Answer
Expert verified
The center is (4, 3) and the radius is 2.
Step by step solution
01
Rearrange the Equation
The given equation of the circle is \(x^2 + y^2 - 8x - 6y + 21 = 0\). To find the center and radius, we need to rearrange the equation in the standard form \((x-h)^2 + (y-k)^2 = r^2\). Start by rearranging the terms: \(x^2 - 8x + y^2 - 6y = -21\).
02
Complete the Square for x
To complete the square for the \(x\) terms: 1. Take the \(x\) terms, \(x^2 - 8x\).2. Find the term to complete the square: \((-8/2)^2 = 16\).3. Rewrite: \(x^2 - 8x = (x-4)^2 - 16\).
03
Complete the Square for y
To complete the square for the \(y\) terms: 1. Take the \(y\) terms, \(y^2 - 6y\).2. Find the term to complete the square: \((-6/2)^2 = 9\).3. Rewrite: \(y^2 - 6y = (y-3)^2 - 9\).
04
Substitute Back and Simplify
Substitute the completed squares back into the equation: \((x-4)^2 - 16 + (y-3)^2 - 9 = -21\).Simplify by combining the constant terms:\((x-4)^2 + (y-3)^2 = -21 + 16 + 9\).Which simplifies to \((x-4)^2 + (y-3)^2 = 4\).
05
Identify the Center and Radius
Compare the equation \((x-4)^2 + (y-3)^2 = 4\) with the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\):- \(h = 4\), \(k = 3\), so the center is \((4, 3)\).- \(r^2 = 4\), so \(r = 2\).Thus, the center of the circle is \((4, 3)\) and the radius is 2.
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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 in algebra to transform a quadratic equation into a form that is easier to work with. It allows us to rewrite a binomial squared expression, like \(ax^2 + bx + c\), in the form \((x-h)^2 + k\). This helps with graphing parabolas and circles.
To complete the square for a quadratic, follow these steps:
To complete the square for a quadratic, follow these steps:
- Identify the quadratic terms: for instance, \(x^2 - 8x\).
- Find the number needed to complete the square by taking half of the linear coefficient \(b\) (which is \(-8\) for our \(x\) terms), square it, and add into the equation. This is calculated as \((-8/2)^2 = 16\).
- Rewrite the quadratic terms as a perfect square by adding and subtracting this number within the expression: \(x^2 - 8x = (x-4)^2 - 16\).
Standard Form of a Circle
The standard form of a circle's equation is very important for determining a circle's attributes, like its center and radius. The equation is given by \((x-h)^2 + (y-k)^2 = r^2\). Here:
When you convert circle equations into standard form, you make it easier to visualize and analyze the circle's location in the coordinate plane.
- \((h, k)\) is the center of the circle.
- \(r\) is the radius of the circle.
When you convert circle equations into standard form, you make it easier to visualize and analyze the circle's location in the coordinate plane.
Center of a Circle
The center of a circle in standard form \((x-h)^2 + (y-k)^2 = r^2\) is simply the point \((h, k)\). This point represents the exact center location of the circle within the coordinate plane.
- For our exercise, comparing the rewritten equation \((x-4)^2 + (y-3)^2 = 4\), we identify that \(h = 4\) and \(k = 3\).
- Therefore, the center of this circle is located at \((4, 3)\).
Radius of a Circle
The radius of a circle is the distance from the center to any point on the circle's boundary. In the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), \(r^2\) denotes the squared radius.
- From the equation \((x-4)^2 + (y-3)^2 = 4\), we have \(r^2 = 4\).
- Taking the square root gives us \(r = 2\).
