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Chapter 3: Problem 50

Exer. 47-56: Find the center and radius of the circle with the given equation. $$ x^{2}+y^{2}-10 x+18=0 $$

Short Answer

Expert verified
Center: (5, 0); Radius: \(\sqrt{7}\).

Step by step solution

01

Group and Reorganize Terms

Start by rearranging the equation \(x^2 + y^2 - 10x + 18 = 0\) to group similar terms together. Focus on the \(x\) terms separately. We have: \(x^2 - 10x + y^2 + 18 = 0\).
02

Complete the Square for x Terms

To complete the square for \(x\) terms, take the coefficient of \(x\), divide by 2, and square it. The coefficient of \(x\) is -10, so we have \((-10/2)^2 = 25\). Add and subtract 25 to complete the square: \(x^2 - 10x + 25 - 25 + y^2 + 18 = 0\).
03

Rewrite as a Perfect Square

Rewrite \(x^2 - 10x + 25\) as a perfect square: \((x - 5)^2\). The equation now looks like \((x - 5)^2 - 25 + y^2 + 18 = 0\).
04

Simplify the Equation

Combine the constants: \(-25 + 18 = -7\). The equation becomes \((x - 5)^2 + y^2 - 7 = 0\).
05

Isolate Terms

Move the numerical part to the other side to isolate the squared terms: \((x - 5)^2 + y^2 = 7\).
06

Identify the Center and Radius

The standard circle equation form is \((x - h)^2 + (y - k)^2 = r^2\). Here \(h = 5\), \(k = 0\), and \(r^2 = 7\). Thus, the center is \((5, 0)\) and the radius is \(\sqrt{7}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
When faced with an equation like \(x^2 + y^2 - 10x + 18 = 0\), the goal is to rewrite it in a form that reveals the circle's center and radius. This involves completing the square, a technique to repackage quadratic expressions into perfect squares.
Here's how you complete the square:
  • Identify the quadratic terms to focus on, in this case, the \(x\) part: \(x^2 - 10x\).
  • Take the coefficient of \(x\), which is -10, divide it by 2 to get -5, and then square it, resulting in 25.
  • Add and subtract this number inside the equation to balance it: \(x^2 - 10x + 25 - 25\).
The equation so far is balanced, but using \((x - 5)^2\) makes all the difference. This transformed expression is easier to manage and gets us closer to a neat circle equation.
Center of Circle
Finding the center of a circle from its equation is essential in geometry. Once our equation \((x - 5)^2 + y^2 = 7\) is structured in the standard form \((x - h)^2 + (y - k)^2 = r^2\), it unveils key circle properties.
  • The terms \((x - h)^2\) and \((y - k)^2\) indicate shifting from the origin \((0, 0)\) to the coordinates \(h, k\).
  • Here, \(h\) and \(k\) are derived directly: \(h = 5\) and \(k = 0\).
In this case, the center is found at the point \((5, 0)\). This point is the center of the circle, defining its location in the coordinate plane and helping understand its positioning.
Circle Radius
The final step in our process involves determining the circle's radius from the equation. Once in standard form, such as \((x - 5)^2 + y^2 = 7\), calculating the radius is straightforward.
  • The right-hand side of the equation represents \(r^2\), where \(r\) is the circle's radius.
  • From \(r^2 = 7\), take the square root of 7 for the actual radius, which is \(\sqrt{7}\).
The radius is a crucial measurement, giving insight into the size of the circle. With a radius of \(\sqrt{7}\), the circle's size and area can be further understood or calculated if necessary.

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Most popular questions from this chapter

Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing, or reflecting.) $$ f(x)=(c x)^{3}+1 ; \quad c=-1,1,4 $$

Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing, or reflecting.) $$ f(x)=-x^{2}+c ; \quad c=-4,2,4 $$

Exer. 53-60: Find a composite function form for \(y\). $$ y=4+\sqrt{x^{2}+1} $$

Exer. 11-20: Find (a) \((f \circ g)(x)\) (b) \((g \circ f)(x)\) (c) \(f(g(-2))\) (d) \(g(f(3))\) $$ f(x)=x^{3}+2 x^{2}, \quad g(x)=3 x $$

Exer. 11-20: Find (a) \((f \circ g)(x)\) (b) \((g \circ f)(x)\) (c) \(f(g(-2))\) (d) \(g(f(3))\) $$ f(x)=|x|, \quad g(x)=-7 $$
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