Chapter 13.1, Problem 12CE

a.

To determine

To find the equation of the circle

Answer to Problem 12CE

The equation of the circle are ${\left(x-2\right)}^2+{\left(y-5\right)}^2={\left(3\right)}^2$

Explanation of Solution

The given center and radius is

  $\text{Center }\left(2,5\right);\text{ radius}=\text{3}$

Calculation:

The equation of the circle can be obtained as

The standard form of the equation of a circle with center and radius is

  ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Where $\left(a,b\right)$ is the center and r is the radius

So, the center $\left(a,b\right)=\left(2,5\right)\text{ and radius }\left(r\right)=\left(3\right)$

Put the values of $a=2,b=5\text{ and }r=3$ on the standard form of the equation

  ${\left(x-2\right)}^2+{\left(y-5\right)}^2={\left(3\right)}^2$

Hence,

The equation of the circle are ${\left(x-2\right)}^2+{\left(y-5\right)}^2={\left(3\right)}^2$

b.

To determine

To find the equation of the circle

Answer to Problem 12CE

The equation of the circle are ${\left(x+2\right)}^2+y^2={\left(5\right)}^2$

Explanation of Solution

The given center and radius is

  $\text{Center }\left(-2,0\right);\text{ radius}=5$

Calculation:

The equation of the circle can be obtained as

The standard form of the equation of a circle with center and radius is

  ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Where $\left(a,b\right)$ is the center and r is the radius

So, the center $\left(a,b\right)=\left(-2,0\right)\text{ and radius }\left(r\right)=\left(5\right)$

Put the values of $a=-2,b=0\text{ and }r=5$ on the standard form of the equation

  ${\left(x+2\right)}^2+y^2={\left(5\right)}^2$

Hence,

The equation of the circle are ${\left(x+2\right)}^2+y^2={\left(5\right)}^2$

c.

To determine

To find the equation of the circle

Answer to Problem 12CE

The equation of the circle are ${\left(x+2\right)}^2+{\left(y-3\right)}^2={\left(10\right)}^2$

Explanation of Solution

The given center and radius is

  $\text{Center }\left(-2,3\right);\text{ radius}=10$

Calculation:

The equation of the circle can be obtained as

The standard form of the equation of a circle with center and radius is

  ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Where $\left(a,b\right)$ is the center and r is the radius

So, the center $\left(a,b\right)=\left(-2,3\right)\text{ and radius }\left(r\right)=\left(10\right)$

Put the values of $a=-2,b=3\text{ and }r=10$ on the standard form of the equation

  ${\left(x+2\right)}^2+{\left(y-3\right)}^2={\left(10\right)}^2$

Hence,

The equation of the circle are ${\left(x+2\right)}^2+{\left(y-3\right)}^2={\left(10\right)}^2$

d.

To determine

To find the equation of the circle

Answer to Problem 12CE

The equation of the circle are ${\left(x-j\right)}^2+{\left(y-k\right)}^2={\left(n\right)}^2$

Explanation of Solution

The given center and radius is

  $\text{Center }\left(j,k\right);\text{ radius}=n$

Calculation:

The equation of the circle can be obtained as

The standard form of the equation of a circle with center and radius is

  ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Where $\left(a,b\right)$ is the center and r is the radius

So, the center $\left(a,b\right)=\left(j,k\right)\text{ and radius }\left(r\right)=\left(n\right)$

Put the values of $a=j,b=k\text{ and }r=n$ on the standard form of the equation

  ${\left(x-j\right)}^2+{\left(y-k\right)}^2={\left(n\right)}^2$

Hence,

The equation of the circle are ${\left(x-j\right)}^2+{\left(y-k\right)}^2={\left(n\right)}^2$