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Chapter 10: Problem 22

For each equation, find the center and radius of the circle. $$ (x+3)^{2}+(y-5)^{2}=81 $$

Short Answer

Expert verified
The center of the circle is (-3,5) and the radius is 9.

Step by step solution

01

Identifying the Parts of the Equation

The equation provided is \((x+3)^{2}+(y-5)^{2}=81\). It is recognizable as the standard form of a circle equation. Here, h is subtracted from x and k is subtracted from y. This means flipping the sign of the x and y components to find the center coordinates. And r is the square root of the number on the right side of the equation for finding the radius.
02

Finding the Center

The components attached to x and y are +3 and -5 respectively. They are subtracted in the equation \( (x-h)^{2} + (y-k)^{2}=r^{2} \), so we flip the signs to get our h and k values. This means our center coordinates, (h,k), are (-3,5).
03

Finding the Radius

The number on the right side of the equation is 81, which represents \(r^{2} \). To find r, we need to take the square root of 81. This gives us a radius of 9.

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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 in geometry represents the point from which every point on the circle's circumference is equidistant. In the standard form of a circle, which is \[ (x-h)^2 + (y-k)^2 = r^2 \], \(h\) and \(k\) denote the \(x\)-coordinate and \(y\)-coordinate of the center, respectively.

To find the center from this form, observe the values that are subtracted from \(x\) and \(y\) in the equation:
  • If the term is \((x+3)^2\), this effectively represents \((x-(-3))^2\), telling us that \(h = -3\).
  • Similarly, \((y-5)^2\) implies \(k = 5\).
Thus, for the equation \((x+3)^2+(y-5)^2=81\), the center of the circle is \((-3, 5)\).

Remember, the signs of \(h\) and \(k\) in the equation are opposite to those in the formula.
Radius of a Circle
The radius of a circle is the distance from the center of the circle to any point on its circumference. In the standard form equation \[ (x-h)^2 + (y-k)^2 = r^2 \], \(r^2\) appears on the right side of the equation.

In order to find the radius (\(r\)) of the circle, you must take the square root of \(r^2\):
  • In our given equation \((x+3)^2+(y-5)^2=81\), the right side of the equation is 81.
  • Thus, \(r^2 = 81\).
  • Taking the square root gives \(r = \sqrt{81} = 9\).
So, the radius of the circle is 9.

It's important to note that \(r\) is always positive as it represents a distance.
Standard Form of a Circle
The standard form of a circle is a way to write the equation that easily shows the center and radius. The standard form is\[ (x-h)^2 + (y-k)^2 = r^2 \],where \( (h, k) \) are the coordinates of the center of the circle, and \( r \) is the radius.

This form allows you to quickly identify:
  • The center of the circle by examining what is added or subtracted directly from the \(x\) and \(y\) variables.
  • The radius by looking at the right-hand side value, which is \(r^2\), and taking its square root.
For example, in the equation \((x+3)^2+(y-5)^2=81\):

  • You can immediately see that the center \((h, k)\) is \((-3, 5)\).
  • The radius \(r\) is \(\sqrt{81} = 9\).
This standard form makes it easy to understand and solve problems related to circles.
This is why recognizing the equation format is so helpful in geometry.

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