Chapter 13.6, Problem 1PSA

(a)

To determine

To find: The equation for the circle from the given data.

Answer to Problem 1PSA

The equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Explanation of Solution

Given:

The centre of the circle is $\left(0,0\right)$ and the radius is $4$ .

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.

Thus, the equation of the circle is,

  $\begin{array}{l}{\left(x-0\right)}^2+\left(y-0^2\right)=4^2\\ x^2+y^2=16\end{array}$

(b)

To determine

To find: The equation for the circle from the given data.

Answer to Problem 1PSA

The equation of the circle is $x^2+{\left(y+2\right)}^2=12$ .

Explanation of Solution

Given:

The centre of the circle is $\left(0,-2\right)$ and the radius is $2\sqrt{3}$ .

Calculation:

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

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

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

Thus, the equation of the circle is,

  $\begin{array}{l}{\left(x-0\right)}^2+{\left(y-\left(-2\right)\right)}^2={\left(2\sqrt{3}\right)}^2\\ x^2+{\left(y+2\right)}^2=12\end{array}$

(c)

To determine

To find: The equation for the circle from the given data.

Answer to Problem 1PSA

The equation of the circle is ${\left(x+4\right)}^2+{\left(y-1\right)}^2=25$ .

Explanation of Solution

Given:

The centre of the circle is $\left(-2,1\right)$ and the radius is $5$ .

Calculation:

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

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

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

Thus, the equation of the circle is,

  $\begin{array}{l}{\left(x-\left(-2\right)\right)}^2+{\left(y-1\right)}^2={\left(5\right)}^2\\ {\left(x+4\right)}^2+{\left(y-1\right)}^2=25\end{array}$

(d)

To determine

To find: The equation for the circle from the given data.

Answer to Problem 1PSA

The equation of the circle is ${\left(x+6\right)}^2+y^2=\frac{1}{4}$ .

Explanation of Solution

Given:

The centre of the circle is $\left(-6,0\right)$ and the radius is $\frac{1}{2}$ .

Calculation:

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

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

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

Thus, the equation of the circle is,

  $\begin{array}{l}{\left(x-\left(-6\right)\right)}^2+{\left(y-0\right)}^2={\left(\frac{1}{2}\right)}^2\\ {\left(x+6\right)}^2+y^2=\frac{1}{4}\end{array}$