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Magazine Chapter 2: Problem 50
Find the center and radius of the circle with the given equation. $$x^{2}+y^{2}+8 x-10 y+37=0$$
Short Answer
Expert verified
The center is (-4, 5) and the radius is 2.
Step by step solution
01
Identify the Standard Form
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. We need to rearrange the given equation in this form.
02
Group x and y Terms
Write the equation as \(x^2 + 8x + y^2 - 10y + 37 = 0\). Notice that the \(x\) terms and \(y\) terms are grouped together.
03
Complete the Square for x Terms
For the \(x\) terms, complete the square: \(x^2 + 8x\) becomes \((x+4)^2 - 16\). Here, take half of 8, which is 4, square it to get 16, and add and subtract 16 to complete the square.
04
Complete the Square for y Terms
For the \(y\) terms, complete the square: \(y^2 - 10y\) becomes \((y-5)^2 - 25\). Take half of -10, which is -5, square it to get 25, and add and subtract 25 to complete the square.
05
Rewrite the Equation
Substitute the completed squares back into the original equation: \((x+4)^2 - 16 + (y-5)^2 - 25 + 37 = 0\).
06
Simplify and Solve for Standard Form
Combine constants: \(-16 - 25 + 37 = -4\). Thus the equation becomes \((x+4)^2 + (y-5)^2 = 4\).
07
Identify Center and Radius
Compare with the standard form \((x-h)^2 + (y-k)^2 = r^2\). The center \((h, k)\) is \((-4, 5)\), and \(r^2 = 4\), so the radius \(r\) is \(2\).
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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 the equation of a circle is a helpful tool for easily identifying a circle's center and radius. It is expressed as \((x-h)^2 + (y-k)^2 = r^2\). Here,
In the given problem, by manipulating the initial equation \(x^2+y^2+8x-10y+37=0\), we can transform it into this standard form to understand more about the circle.
- \((h, k)\) represents the center of the circle.
- \(r\) is the radius of the circle.
In the given problem, by manipulating the initial equation \(x^2+y^2+8x-10y+37=0\), we can transform it into this standard form to understand more about the circle.
Completing the Square
Completing the square is a technique that helps transform a quadratic equation into a more useful format, especially when seeking the standard form of a circle. It involves turning a binomial into a perfect square trinomial.
To complete the square:
To complete the square:
- First, focus on the \(x\) terms: \(x^2 + 8x\). Take half of 8, which is 4, and square it to get 16. This changes \(x^2 + 8x\) to \((x+4)^2 - 16\).
- Next, address the \(y\) terms: \(y^2 - 10y\). Take half of -10, which is -5, square it to get 25, and rewrite \(y^2 - 10y\) as \((y-5)^2 - 25\).
Center of a Circle
The center of a circle is a pivotal concept in geometry, acting as a reference point for the circle's existence in the plane. From the standard form, \((x-h)^2 + (y-k)^2 = r^2\), the center is easy to spot as \((h, k)\).
In the equation obtained after completing the square, \((x + 4)^2 + (y-5)^2 = 4\), we substitute and read off the values of \(h\) and \(k\). By comparing, \((h, k) = (-4, 5)\) is identified.
In the equation obtained after completing the square, \((x + 4)^2 + (y-5)^2 = 4\), we substitute and read off the values of \(h\) and \(k\). By comparing, \((h, k) = (-4, 5)\) is identified.
- Here, \(-4\) is derived from \((x+4)\), as it's transformed to \(x - (-4)\).
- Similarly, 5 is derived from \((y - 5)\), maintaining its position.
Radius of a Circle
The radius of a circle is a measure of the constant distance from the center to any point on the circle. It can be easily identified from the standard circle equation form. In the equation \((x-h)^2 + (y-k)^2 = r^2\), the term \(r^2\) gives the square of the radius.
For our reformed equation \((x+4)^2 + (y-5)^2 = 4\), we identify \(r^2 = 4\).
For our reformed equation \((x+4)^2 + (y-5)^2 = 4\), we identify \(r^2 = 4\).
- The radius \(r\), which is the square root of \(r^2\), is \(2\) (since \(\sqrt{4} = 2\)).
