Chapter 13.6, Problem 11PSB

(a)

To determine

To find: The centre and the radius of the circle represented by the given equation.

Answer to Problem 11PSB

The centre of the circle is $\left(h,k\right)=\left(0,4\right)$ and the radius is $5$ .

Explanation of Solution

Given:

The given equation is $x^2+y^2-8y=9$ .

Calculation:

Consider the given equation is,

  $x^2+y^2-8y=9$

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.

Convert given equation in the general form as,

  $\begin{array}{l}{x}^2+y^2-8y+2\left(8\right)=9+2\left(8\right)\\ x^2+y^2-8y+16=9+16\\ x^2+{\left(y-4\right)}^2=25\\ x^2+{\left(y-4\right)}^2=5^2\end{array}$

Thus, the centre of the circle is $\left(h,k\right)=\left(0,4\right)$ and the radius is $5$ .

(b)

To determine

To find: The centre and the radius of the circle represented by the given equation.

Answer to Problem 11PSB

The centre of the circle is $\left(h,k\right)=\left(-7,-3\right)$ and the radius is $6$ .

Explanation of Solution

Given:

The given equation is ${\left(x+7\right)}^2+y^2+6y=27$ .

Calculation:

Consider the given equation is,

  ${\left(x+7\right)}^2+y^2+6y=27$

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.

Convert given equation in the general form as,

  $\begin{array}{l}{\left(x+7\right)}^2+y^2+6y+9=27+9\\ {\left(x+7\right)}^2+{\left(y+3\right)}^2=36\\ {\left(x+7\right)}^2+{\left(y+3\right)}^2=6^2\end{array}$

Thus, the centre of the circle is $\left(h,k\right)=\left(-7,-3\right)$ and the radius is $6$ .

(c)

To determine

To find: The centre and the radius of the circle represented by the given equation.

Answer to Problem 11PSB

The centre of the circle is $\left(h,k\right)=\left(-5,6\right)$ and the radius is $\sqrt{51}$ .

Explanation of Solution

Given:

The given equation is $x^2+10x+y^2-12y=-10$ .

Calculation:

Consider the given equation is,

  $x^2+10x+y^2-12y=-10$

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.

Convert given equation in the general form as,

  $\begin{array}{l}{x}^2+10x+25+y^2-12y+36=-10+36+25\\ {\left(x+5\right)}^2+{\left(y-6\right)}^2={\left(\sqrt{51}\right)}^2\end{array}$

Thus, the centre of the circle is $\left(h,k\right)=\left(-5,6\right)$ and the radius is $\sqrt{51}$ .

(d)

To determine

To find: The centre and the radius of the circle represented by the given equation.

Answer to Problem 11PSB

The centre of the circle is $\left(h,k\right)=\left(4,-7\right)$ and the radius is $10$ .

Explanation of Solution

Given:

The given equation is $x^2+y^2=8x-14y+35$ .

Calculation:

Consider the given equation is,

  $\begin{array}{l}{x}^2+y^2=8x-14y+35\\ x^2+y^2-8x+14y=35\end{array}$

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.

Convert given equation in the general form as,

  $\begin{array}{l}{x}^2+y^2-8x+14y+16+49=35+16+49\\ {\left(x-4\right)}^2+{\left(y+7\right)}^2={10}^2\end{array}$

Thus, the centre of the circle is $\left(h,k\right)=\left(4,-7\right)$ and the radius is $10$ .