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Chapter 10: Problem 10

Find the center and radius of the circle whose equation is given. $$x^{2}+y^{2}+10 x-75=0$$

Short Answer

Expert verified
Answer: The center of the circle is \((-5, 0)\) and the radius is \(10\).

Step by step solution

01

Rewrite the given equation

Start by rewriting the given equation as follows: $$x^{2}+10 x+y^{2}=-75$$
02

Complete the square for x terms

Write the linear term's coefficient of x, which is 10, in the form of 2ax. We have 2ax = 10, so a = 5. Now, we need to add and subtract \((a)^2\) = \((5)^2\) = 25 to complete the square. $$x^{2}+10 x+25+y^{2}=-75+25$$
03

Complete the square for y terms

Since there is no linear term for y, we do not need to complete the square for y terms. Also, arrange the terms to have the standard form of a circle's equation. $$(x+5)^2+y^2=100$$
04

Determine the center and the radius

Compare the equation, \((x+5)^2+y^2=100\) , with the standard form of a circle's equation, \((x-a)^2 + (y-b)^2 = r^2\): $$(x-(-5))^2+(y-0)^2=10^2$$ So we can clearly see that the center of the circle, \((a,b)\) is equal to \((-5,0)\) and the radius, \(r\) is equal to \(10\).

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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 mathematical technique used to convert a quadratic equation into a perfect square trinomial, making it easier to solve or manipulate. For a quadratic in the form of \(ax^2 + bx + c\), completing the square involves rearranging it to the form \((x+d)^2 + e\). This method is particularly useful when finding the equation of a circle.
Here's how it works:
  • Take the coefficient of the \(x\) term (here it's 10 from the equation \(x^2+10x+y^2=-75\)).
  • Divide it by 2 and square the result. So, \(10/2 = 5\) and \((5)^2 = 25\).
  • Add and subtract this square to/from the original equation, effectively "completing the square". Thus, the equation becomes \((x+5)^2 + y^2 = 100\).
This procedure transforms the expression into a perfect square form, aiding in identifying features like the center and the radius of the circle.
Center of a Circle
The center of a circle represented by the equation \( (x-a)^2 + (y-b)^2 = r^2 \) can be found directly from the variables \(a\) and \(b\). In a circle's equation in standard form, these represent horizontal and vertical shifts from the origin.
  • After completing the square, the equation \((x+5)^2 + y^2 = 100\) reveals the center.
  • The inner part \((x-a)^2\) is \((x+5)^2\), implying \(a = -5\).
  • For the \(y\) part, \((y-b)^2\) simplifies to \((y-0)^2\), meaning \(b = 0\).
Therefore, the circle's center is located at \((-5, 0)\). This represents the circle's precise point pivot for all rotations.
Radius of a Circle
The radius of a circle is a measure of length from its center to any point on its boundary. This measurement is typically expressed as \(r\) in the standard equation of a circle's format: \( (x-a)^2 + (y-b)^2 = r^2 \).
  • Upon completing the square to reform the equation to \((x+5)^2 + y^2 = 100\), the circle's proper radius can be observed.
  • This expression equates to \((x-(-5))^2 + (y-0)^2 = 10^2\), which clearly shows that \(r^2 = 100\).
  • Here, the radius, \(r\), is the square root of \(100\), so \(r = 10\).
Knowing the radius is crucial, as it helps determine the size and the span of the circle from its center.
Standard Form of Circle Equation
The standard form of a circle's equation is a way of expressing the circle by emphasizing its center and radius, making calculations involving circles more approachable. It takes the form \((x-a)^2 + (y-b)^2 = r^2\), where \(a\) and \(b\) are the coordinates of the center and \(r\) is the radius.
  • The equation \(x^2+10x+y^2 = -75\) was put into standard form using completing the square
  • This resulted in \((x+5)^2 + y^2 = 100\), where \((-5, 0)\) is the center and \(10\) is the radius.
  • The equation describes all the points \( (x, y) \) that lie perfectly on the circle represented.
Understanding the standard form helps in directly recognizing crucial geometric properties such as positioning and scale, making it valuable in graphing and further analysis of circular equations.

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