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Chapter 0: Problem 19

Find the center and radius of the circle with the given equation. $$ x^{2}+y^{2}-12 x+35=0 $$

Short Answer

Expert verified
Center: (6, 0); Radius: 1.

Step by step solution

01

Rearrange the Equation

Start by rearranging the given circle equation, \( x^2 + y^2 - 12x + 35 = 0 \), to group the \( x \) terms together with \( x^2 \), and separate out the \( y^2 \) term: \( x^2 - 12x + y^2 = -35 \).
02

Complete the Square for x

To complete the square for the \( x \) terms, take half of the coefficient of \( x \), which is \(-12\). Half of \(-12\) is \(-6\). Then square \(-6\) to get \(36\). Add \(36\) to both sides of the equation to balance it, resulting in \( x^2 - 12x + 36 + y^2 = -35 + 36 \). This simplifies to \((x-6)^2 + y^2 = 1\).
03

Identify the Center and Radius

Now that the equation is in the form \((x-h)^2 + (y-k)^2 = r^2\), identify the center \((h, k)\) and radius \(r\). The expression \((x-6)^2 + y^2 = 1\) indicates that \(h = 6\), \(k = 0\), and \(r^2 = 1\). Thus, the center of the circle is \((6, 0)\) and the radius is \(r = \sqrt{1} = 1\).

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Key Concepts

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

Completing the Square
When dealing with the equation of a circle, often times it is not given in its standard form. To transform a quadratic equation into a recognizable circle equation, one of the most effective techniques used is **completing the square**.
This method helps to rewrite the equation into a perfect square form, making it easier to identify properties of the circle such as the center and the radius.
Here's a quick step-by-step guide:
  • Look at the terms involving the variable, such as the quadratic and linear terms. For a monomial like \( x^2 - 12x \), you'll focus on the \( x \)-related terms.
  • Take half of the linear coefficient (in this case, for \(-12x\) it is \(-6\)), and then square it to make a complete square, which gives you \(36\).
  • Add and subtract the squared term in the equation to balance it. This ensures the equation remains equivalent while transforming it into a form that can be easily manipulated.
Once completed, this process rewrites the equation into a readable format, specifically helpful for pinpointing the circle's center and radius.
Circle Center
After completing the square, the equation of the circle generally appears as \((x-h)^2 + (y-k)^2 = r^2\). From this, the center of the circle \((h, k)\) is straightforward to identify.
The term \((x-h)^2\) indicates a shift of the circle's center \(h\) units along the x-axis, while the \(k\) in \((y-k)^2\) represents a vertical movement along the y-axis.
In our rearranged equation \((x-6)^2 + y^2 = 1\), you can discern:
  • For the x-component, \(h=6\) showing a shift to the right.
  • For the y-component, \(k=0\), meaning no vertical shift has occurred.
Thus, the center of the circle is plotted at \((6, 0)\) on a coordinate plane. Recognizing the circle's center is crucial for graphing and understanding its geometry intuitively.
Circle Radius
Finally, moving on to calculating the **circle radius**, we use the transformed circle equation, \((x-h)^2 + (y-k)^2 = r^2\). Here, \(r^2\) is the squared value of the circle's radius.
The provided example simplifies to \((x-6)^2 + y^2 = 1\), with \(r^2 = 1\). Thus, the radius \(r\) is derived by taking the square root of \(r^2\).
  • The calculation \(r = \sqrt{1} = 1\) tells us the radius is a single unit long.
The radius is essential not only for comprehending the scale of the circle on a graph but also for calculations involving the circumference and area of the circle. Knowing just how wide the circle spans from its center can influence our understanding of various mathematical concepts related to circles.

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