Chapter 13.6, Problem 3PSA

(a)

To determine

To find: The centre, the radius, the diameter, the circumference and the area of the circle for the given equation.

Answer to Problem 3PSA

The centre of the circle is at $\left(0,0\right)$ , the radius is 6, the diameter is 12, circumference is $12\pi$ and the area is $36\pi$ .

Explanation of Solution

Given:

The given equation is $x^2+y^2=36$ .

Calculation:

Consider the formula for the equation of the circle is of the form as,

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

Here, $\left(h,k\right)$ is the centre and $r$ is the radius.

Consider the given equation of the circle is,

  $x^2+y^2=36$

From the given equation and the general equation the centre of the circle is,

  $\left(h,k\right)=\left(0,0\right)$

The radius of the circle is,

  $\begin{array}{l}{r}^2=36\\ r=6\end{array}$

The diameter of the circle is,

  $\begin{array}{c}d=2\left(6\right)\\ =12\end{array}$

The circumference of the circle is,

  $\begin{array}{c}C=\pi d\\ =12\pi \end{array}$

The area of the circle is obtained as,

  $\begin{array}{c}A=\pi r^2\\ =\pi {\left(6\right)}^2\\ =36\pi \end{array}$

Thus, the centre of the circle is at $\left(0,0\right)$ , the radius is 6, the diameter is 12, circumference is $12\pi$ and the area is $36\pi$ .

(b)

To determine

To find: The centre, the radius, the diameter, the circumference and the area of the circle for the given equation.

Answer to Problem 3PSA

The centre of the circle is at $\left(3,6\right)$ , the radius is $10$ , the diameter is $20$ , circumference is $20\pi$ and the area is $100\pi$ .

Explanation of Solution

Given:

The given equation is ${\left(x-3\right)}^2+{\left(y+6\right)}^2=100$ .

Calculation:

Consider the formula for the equation of the circle is of the form as,

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

Here, $\left(h,k\right)$ is the centre and $r$ is the radius.

Consider the given equation of the circle is,

  ${\left(x-3\right)}^2+{\left(y+6\right)}^2=100$

From the given equation and the general equation the centre of the circle is,

  $\left(h,k\right)=\left(3,6\right)$

The radius of the circle is,

  $\begin{array}{l}{r}^2=100\\ r=10\end{array}$

The diameter of the circle is,

  $\begin{array}{c}d=2\left(10\right)\\ =20\end{array}$

The circumference of the circle is,

  $\begin{array}{c}C=d\pi \\ =20\pi \end{array}$

The area of the circle is obtained as,

  $\begin{array}{c}A=\pi {\left(r\right)}^2\\ =\pi {\left(10\right)}^2\\ =100\pi \end{array}$

Thus, the centre of the circle is at $\left(3,6\right)$ , the radius is $10$ , the diameter is $20$ , circumference is $20\pi$ and the area is $100\pi$ .

(c)

To determine

To find: The centre, the radius, the diameter, the circumference and the area of the circle for the given equation.

Answer to Problem 3PSA

The centre of the circle is at $\left(-5,0\right)$ , the radius is $\frac{3}{2}$ , the diameter is $3$ , circumference is $3\pi$ and the area is $\frac{9}{4}\pi$ .

Explanation of Solution

Given:

The given equation is ${\left(x+5\right)}^2+{\left(y\right)}^2=\frac{9}{4}$ .

Calculation:

Consider the formula for the equation of the circle is of the form as,

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

Here, $\left(h,k\right)$ is the centre and $r$ is the radius.

Consider the given equation of the circle is,

  ${\left(x+5\right)}^2+{\left(y\right)}^2=\frac{9}{4}$

From the given equation and the general equation the centre of the circle is,

  $\left(h,k\right)=\left(-5,0\right)$

The radius of the circle is,

  $\begin{array}{l}{r}^2=\frac{9}{4}\\ r=\frac{3}{2}\end{array}$

The diameter of the circle is,

  $\begin{array}{c}d=2\left(\frac{3}{2}\right)\\ =3\end{array}$

The circumference of the circle is,

  $\begin{array}{c}C=d\pi \\ =3\pi \end{array}$

The area of the circle is obtained as,

  $\begin{array}{c}A=\pi {\left(r\right)}^2\\ =\pi {\left(\frac{3}{2}\right)}^2\\ =\frac{9}{4}\pi \end{array}$

Thus, the centre of the circle is at $\left(-5,0\right)$ , the radius is $\frac{3}{2}$ , the diameter is $3$ , circumference is $3\pi$ and the area is $\frac{9}{4}\pi$ .

(d)

To determine

To find: The centre, the radius, the diameter, the circumference and the area of the circle for the given equation.

Answer to Problem 3PSA

The centre of the circle is at $\left(-5,2\right)$ , the radius is $9$ , the diameter is $18$ , circumference is $18\pi$ and the area is $81\pi$ .

Explanation of Solution

Given:

The given equation is $\frac{{\left(x+5\right)}^2}{3}+{\frac{\left(y-2\right)}{3}}^2=27$ .

Calculation:

Consider the formula for the equation of the circle is of the form as,

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

Here, $\left(h,k\right)$ is the centre and $r$ is the radius.

Consider the given equation of the circle is,

  $\begin{array}{l}\frac{{\left(x+5\right)}^2}{3}+{\frac{\left(y-2\right)}{3}}^2=27\\ {\left(x+5\right)}^2+\left(y-2\right)=27\times 3\\ {\left(x+5\right)}^2+{\left(y-2\right)}^2=81\end{array}$

From the given equation and the general equation the centre of the circle is,

  $\left(h,k\right)=\left(-5,2\right)$

The radius of the circle is,

  $\begin{array}{l}{r}^2=81\\ r=9\end{array}$

The diameter of the circle is,

  $\begin{array}{c}d=2\left(9\right)\\ =18\end{array}$

The circumference of the circle is,

  $\begin{array}{c}C=18\pi \\ =18\pi \end{array}$

The area of the circle is obtained as,

  $\begin{array}{c}A=\pi {\left(r\right)}^2\\ =\pi {\left(9\right)}^2\\ =81\pi \end{array}$

Thus, the centre of the circle is at $\left(-5,2\right)$ , the radius is $9$ , the diameter is $18$ , circumference is $18\pi$ and the area is $81\pi$ .