Chapter 9.4, Problem 16MCQ

To determine

To find: What is the radius of the circle.

Answer to Problem 16MCQ

The radius of the circle is $r=7$ .

Explanation of Solution

Given information:

The given equation is,

  $x^2+2x+y^2+14y+34=0$

The given equation is

  $x^2+2x+y^2+14y+34=0$

Add and subtract $1$ to left side of the equation

  $\begin{array}{l}{x}^2+2x+1-1+y^2+14y+34=0\\ {\left(x+1\right)}^2+y^2+14y+34-1=0\\ {\left(x+1\right)}^2+y^2+14y+33=0\mathrm{...}\left(1\right)\end{array}$

Add and subtract $49$ to left side of the equation $\mathrm{...}\left(1\right)$

  $\begin{array}{l}{\left(x+1\right)}^2+y^2+14y+49-49+33=0\\ {\left(x+1\right)}^2+{\left(y+7\right)}^2-49+33=0\\ {\left(x+1\right)}^2+{\left(y+7\right)}^2-16=0\\ {\left(x+1\right)}^2+{\left(y+7\right)}^2=16\mathrm{...}\left(2\right)\end{array}$

The standard form of a circle is

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{...}\left(3\right)$

There are $\left(h,k\right)$ is the centre point.

  $\left(x,y\right)$ is point on the circle and $r$ is the radius of the circle

To compare the equation $\left(2\right)$ from $\left(3\right)$

The radius of the circle is $r=4$

The centre of the circle is $h=-1,k=-7$

Thus, the radius of the circle is $r=4$ .