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Chapter 1: Q 8. (page 50)

Find the center and radius of the circle. Write the standard form of the equation.

Short Answer

Expert verified

Center is $(1, 2)$ .

Radius is equal to $2$ units.

Standard form of the equation is $(x - 1)^{2} + (y - 2)^{2} = 4$ .

Step by step solution

01

Step 1. Given information

Figure of a circle is as follows:

02

Step 2. Find the center and radius of a circle.

Radius is the distance between the center and any point on the circle

In this case, radius is the distance between the points $(1, 2)$ and $(1, 0)$ .

Radius $= \sqrt{(1 - 1)^{2} + (2 - 0)^{2}}$

$= \sqrt{0 + 4} = \sqrt{4} = 2$

From the given graph, we can see that the center is $(1, 2)$ .

03

Step 3. Find the standard form of the equation.

Equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^{2} + (y - k)^{2} = r^{2}$ .

Therefore, equation of the circle with center $(1, 2)$ and radius $2$ is

$\begin{array}{rcl} (x - 1)^{2} + (y - 2)^{2} & = & 2^{2} \\ (x - 1)^{2} + (y - 2)^{2} & = & 4 \end{array}$

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Most popular questions from this chapter

In Problem 17-36, solve each equation algebraically. Verify your solution using a graphing utility.

$x^{2} - 7 x - 18 = 0$

True or false The point $(1, 2)$ is on the line $2 x + y = 4$ .

In Problem 17-36, solve each equation algebraically. Verify your solution using a graphing utility.

$\frac{5}{y} + \frac{4}{y} = 3$

In Problem 17-36, solve each equation algebraically. Verify your solution using a graphing utility.

$x^{3} + x^{2} - 4 x - 4 = 0$

In Problem 17-36, solve each equation algebraically. Verify your solution using a graphing utility.

$\frac{2}{x + 2} + \frac{3}{x - 1} = - \frac{8}{5}$

See all solutions

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