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Chapter 2: 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

Move constant to the other side

Start by rewriting the equation to separate constants from the variables. The given equation is:\[ x^2 + y^2 - 10x + 18 = 0 \]Move the constant term to the right side:\[ x^2 + y^2 - 10x = -18 \]
02

Complete the square for the x terms

To complete the square for the x terms, we focus on the terms involving x:\[ x^2 - 10x \]Take half of the coefficient of x, which is -10, to get -5, and square it, which gives 25. Add and subtract 25 inside the equation:\[ (x^2 - 10x + 25) - 25 + y^2 = -18 \]
03

Form a perfect square trinomial

Rewrite the expression containing x as a perfect square trinomial:\[ (x - 5)^2 - 25 + y^2 = -18 \]
04

Move new constant terms to the other side

Add 25 to both sides to balance the equation:\[ (x - 5)^2 + y^2 = 7 \]
05

Identify the center and radius

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.By comparing, we see that:- Center (h, k) is (5, 0)- Radius r is \( \sqrt{7} \)
06

Final Verification

Make sure the derived equation \( (x - 5)^2 + y^2 = 7 \) is consistent with a circle in the standard form, confirming the calculations for center and radius.

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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 convert a quadratic expression into a perfect square trinomial. This process is vital for solving quadratic equations and finding the standard form of a circle equation. In the given problem, we start with an equation:
\[x^2 + y^2 - 10x + 18 = 0\]
Our goal is to complete the square for the terms involving \(x\). Follow these simple steps:
  • Focus on the \(x\) terms: \(x^2 - 10x\).
  • Take half of the coefficient of \(x\) (here, \(-10\)), resulting in \(-5\), and then square it. You get 25.
  • Add and subtract this squared value (25) to the expression.
So, it becomes:
\[(x^2 - 10x + 25) - 25 + y^2 = -18\]
This method rearranges the quadratic equation into a more workable form, where one part is a perfect square trinomial.
Standard Form of a Circle
The standard form of a circle is essential in mathematics to easily identify the circle's properties, like its center and radius. This format is:
\[(x - h)^2 + (y - k)^2 = r^2\]
Here, \((h, k)\) is the circle's center, and \(r\) is the radius. After completing the square in the problem, the equation is brought to:
\[(x - 5)^2 + y^2 = 7\]
In this structure, it's easy to spot the values that describe the circle's key characteristics:
  • Like terms from the original equation become recognized parts of a geometric shape.
  • Position (shift of the graph) is clear by the terms \((x-5)\).
This form makes it straightforward to evaluate, draw, and interpret circles in math, enabling students to instantly read off the center and radius.
Center and Radius of a Circle
By understanding the standard form conversion, we can now extract the center and radius from the circle equation. From:
\[(x - 5)^2 + y^2 = 7\]
We can directly see:
  • The center of the circle, represented by \((h, k)\), is found at \((5, 0)\).
  • The radius \(r\) is the square root of the right-hand side constant, \(\sqrt{7}\).
Understanding these components is crucial as they describe the circle's placement and size:
  • The center is the middle point around which the circle is perfectly balanced.
  • The radius determines how far the boundary is from the center in all directions.
Grasping these ideas allows for practical applications in geometry, helping to compute distances and generate circular graphs.

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