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Chapter 2: Problem 52

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

Rearrange the equation

Start by rearranging the given equation to group the terms involving the same variable. Rewrite it as:\[ \left( x^{2} - 10x \right) + y^{2} = -18 \] This allows us to focus on completing the square for the terms involving \(x\).
02

Complete the square for x

To complete the square for \(x^2 - 10x\), take half of the coefficient of \(x\), which is \(-10\), divide by 2 to get \(-5\), and then square it to get \(25\). Add and subtract \(25\) inside the equation:\[ \left( x^{2} - 10x + 25 - 25 \right) + y^{2} = -18 \]This helps to form a perfect square trinomial for \(x\).
03

Simplify the equation

Rewrite the equation from Step 2 by recognizing the perfect square trinomial:\[ \left( x - 5 \right)^{2} - 25 + y^{2} = -18 \]Now add 25 to both sides to balance the equation:\[ \left( x - 5 \right)^{2} + y^{2} = 7 \]
04

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. From the equation \( (x - 5)^2 + y^2 = 7 \), 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 working with quadratic equations, completing the square is a valuable method to rearrange terms into a perfect square trinomial. This technique is especially useful when dealing with circle equations. Completing the square involves:
  • Taking the coefficient of the linear term (in our case, \(-10\) from \(x^2 - 10x\))
  • Dividing this coefficient by 2 to find the midpoint
  • Squaring the result to ensure we can form a perfect square
For example, in the expression \(x^2 - 10x\), we divide \(-10\) by 2 to get \(-5\), and then square it to obtain 25. By adding and subtracting 25, we construct a complete square: \((x - 5)^2\). This eases simplifying and solving the equation to find circle properties.
Center of a Circle
Finding the center of a circle is crucial for understanding its position on the coordinate plane. The center is defined from the circle's standard equation: \((x - h)^2 + (y - k)^2 = r^2\).
The terms \(h\) and \(k\) indicate the horizontal and vertical shifts from the origin, respectively.
In our example, from the equation \((x - 5)^2 + y^2 = 7\), we identify the center as \( (5, 0) \).
  • \((h, k) = (5, 0)\): The circle is shifted 5 units to the right from the origin and remains on the x-axis since the y term wasn't shifted.
Knowing where the circle is located helps in visualizations and further calculations.
Radius of a Circle
The radius of a circle is the distance from its center to any point on the circumference. Having the standard equation format makes identifying this straightforward.
The radius is derived from the equation: \((x - h)^2 + (y - k)^2 = r^2\), where \(r^2\) is the number on the right side.
From \((x - 5)^2 + y^2 = 7\), the radius \(r\) is the square root of 7.
  • Radius \( r = \sqrt{7} \): This is approximately 2.65, giving the length of each line segment from the center \((5, 0)\) to any point on the circle's boundary.
Understanding the radius aids in determining the size of the circle and its spatial extent.
Standard Form of a Circle
The standard form of a circle鈥檚 equation is fundamental in identifying its geometric properties like center and radius. This form is expressed as: \((x - h)^2 + (y - k)^2 = r^2\), providing a clear view of:
  • \((h, k)\): The circle鈥檚 center coordinates.
  • \(r\): The radius length.
For instance, converting the equation \(x^2 + y^2 - 10x + 18 = 0\) into \((x - 5)^2 + y^2 = 7\) showcases the circle鈥檚 center at \( (5, 0) \) and radius \( \sqrt{7} \).
Recognizing the standard form supports solving more complex problems, graphing circles, and translating equations into meaningful geometric insights.

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