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Chapter 12: Problem 11

Write the standard equation for each circle with the given center and radius. Center \((-6,-3),\) radius \(\frac{1}{2}\)

Short Answer

Expert verified
\( (x + 6)^2 + (y + 3)^2 = \frac{1}{4}\)

Step by step solution

01

Identify the given parameters

The center of the circle is given as \((-6, -3)\) and the radius is \(\frac{1}{2}\).
02

Recall the standard form of a circle equation

The standard form of a circle's equation with center \( (h, k) \) and radius \( r \) is: \[ (x - h)^2 + (y - k)^2 = r^2 \]
03

Substitute the given center coordinates and radius

Substitute \(-6\) for \ h \, \(-3\) for \ k \, and \ \frac{1}{2} \ for \ r \ into the standard form equation: \[(x - (-6))^2 + (y - (-3))^2 = \left( \frac{1}{2} \right)^2\]
04

Simplify the equation

Simplify the equation to get the standard form: \[(x + 6)^2 + (y + 3)^2 = \frac{1}{4} \]

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Key Concepts

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

center of a circle
The center of a circle is a crucial point for identifying and working with circles on a coordinate plane. It鈥檚 denoted as \(h, k\) in the standard equation of a circle, which is \[ (x - h)^2 + (y - k)^2 = r^2 \]. The coordinates \(-6, -3\) tell us the horizontal and vertical shift from the origin.
  • If the center is \(-6\), that means it鈥檚 6 units to the left from the origin.
  • If the center is \(-3\), that means it鈥檚 3 units down from the origin.
This is important because any point \((x, y)\) on the circle will maintain an equal distance (the radius) from this center point.
radius of a circle
The radius of a circle is the distance from the center to any point on the circle. It's represented by \(r\) in the equation. Here, our radius is \(\frac{1}{2}\). When you plug the radius into the standard equation, you square it to keep everything non-negative. For example, though the radius is \(\frac{1}{2}\), it becomes \(\frac{1}{4}\) in the equation: \[ (x + 6)^2 + (y + 3)^2 = \frac{1}{4} \]
  • A smaller radius results in a smaller circle.
  • Make sure you always square the radius when substituting into the formula.
simplifying equations
Simplifying equations involves making them as basic as possible while preserving the original relationship. During simplification:
  • Substitute known values into the equation. For instance, \( x - (-6) \) simplifies to \( x + 6 \).
  • Ensure to correctly follow steps, like squaring fractions. Here, \(\frac{1}{2}\) squared becomes \(\frac{1}{4}\).
This step-by-step breakdown makes it easier to understand and solve for any required variable. In the example given, the final simplified equation reads: \[ (x + 6)^2 + (y + 3)^2 = \frac{1}{4} \]

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Most popular questions from this chapter

Sketch the graph of each hyperbola. $$ \frac{(y+3)^{2}}{16}-\frac{(x+1)^{2}}{9}=1 $$

Graph each relation on a graphing calculator by solving for \(y\) and graphing two functions. $$x=(y+2)^{2}-1$$

Determine whether the graph of each equation is a circle, parabola, ellipse, or hyperbola. $$ y=x^{2}+1 $$

Solve each problem. Determine the equation of the circle that is centered at\((2,3)\) and passes through \((-2,-1)\)

Graph both equations of each system on the same coordinate axes. Use elimination of variables to find all points of intersection. $$ \begin{aligned} &y=x^{2}-2\\\ &y=2 x-3 \end{aligned} $$
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