Chapter 1, Problem 18T

(a)

To determine

To find: The center and radius of circle $x^2+y^2=25$ and draw its graph.

Answer to Problem 18T

The center of circle is $\left(0,0\right)$ and radius of circle is 5.

Explanation of Solution

Section1:

The given equation of circle is $x^2+y^2=25$

The equation of circle with center $\left(0,0\right)$ and radius r is,

$x^2+y^2=r^2$. (1)

Rewrite the given equation,

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

Compare above equation with equation (1) to get the center $\left(0,0\right)$ and radius $r=5$ of circle.

Thus, the center of circle is $\left(0,0\right)$ and radius of circle is 5.

Section2:

From the section1 the center of circle is pass through origin and the radius of circle is 5 that is shown below,

Figure (1)

Figure (1) shows the graph of circle $x^2+y^2=25$ with center $\left(0,0\right)$ and radius 5.

(b)

To determine

To find: The center and radius of circle ${\left(x-2\right)}^2+{\left(y+1\right)}^2=9$ and draw its graph.

Answer to Problem 18T

The center of circle is $\left(2,-1\right)$ and radius of circle is 3.

Explanation of Solution

Section1:

The given equation of circle is ${\left(x-2\right)}^2+{\left(y+1\right)}^2=9$

The equation of circle with center $\left(h,k\right)$ and radius r is,

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$. (1)

Rewrite the given equation,

${\left(x-2\right)}^2+{\left(y+1\right)}^2={\left(3\right)}^2$

Compare given equation with above equation to get the center $\left(2,-1\right)$ and radius $r=3$ of circle.

Thus, the center of circle is $\left(2,-1\right)$ and radius of circle is 3.

Section2:

The given equation of circle is ${\left(x-2\right)}^2+{\left(y+1\right)}^2=9$.

From the section1 the center of circle is $\left(2,-1\right)$ and the radius of circle is 3 that is shown below,

Figure (2)

Figure (2) shows the graph of circle ${\left(x-2\right)}^2+{\left(y+1\right)}^2=9$ with center $\left(2,-1\right)$ and radius 3.

(c)

To determine

To find: The center and radius of circle $x^2+6x+y^2-2y+6=0$ and draw its graph.

Answer to Problem 18T

The center of circle is $\left(-3,1\right)$ and radius of circle is 2.

Explanation of Solution

Given:

The equation is $x^2+6x+y^2-2y+6=0$.

Calculation:

Section1:

The equation of circle with center $\left(h,k\right)$ and radius r is,

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$. (1)

First, group the x-terms and y-terms then make a complete square within each group in the given equation,

$\left(x^2+6x\right)+\left(y^2-2y\right)=-6$

Add ${\left(\frac{1}{2}\cdot 6\right)}^2=9$ to complete the square of $\left(x^2+6x\right)$ and ${\left(\frac{1}{2}\cdot \left(-2\right)\right)}^2=1$ to complete the square of $\left(y^2-2y\right)$ and solve for center and radius of circle,

$\begin{array}{c}\left(x^2+6x+9\right)+\left(y^2-2y+1\right)=-6+9+1\\ \left(x^2+2x3+{\left(3\right)}^2\right)+\left(y^2-2y1+{\left(1\right)}^2\right)=4\\ {\left(x+3\right)}^2+{\left(y-1\right)}^2={\left(2\right)}^2\end{array}$

Compare the above equation with equation (1) to get the center $\left(-3,1\right)$ and radius $r=2$ of circle.

Thus, the center of circle is $\left(-3,1\right)$ and radius of circle is 2.

Section2:

From the section1 the center of circle is $\left(-3,1\right)$ and the radius of circle is 2 that is shown below,

Figure (3)

Figure (3) shows the graph of circle $x^2+6x+y^2-2y+6=0$ with center $\left(2,-1\right)$ and radius 2.