Chapter 1, Problem 107RE

To determine

To interpret:The given equation $x^2+y^2+2x-6y+9=0$ and decide if it represents a circle, a point or in the last case if it does not represent any graph.

Answer to Problem 107RE

The given equation $x^2+y^2+2x-6y+9=0$ represents a circle with center $\left(-1,3\right)$ and radius $1$ .

Explanation of Solution

Given information:

The equation $x^2+y^2+2x-6y+9=0$

Formula used:

For a given circle with center $\left(h,k\right)$ and radius $r$ the equation is given as:

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

This is referred to as the Standard form for the equation of a given circle.

Moreover, if the center of the circle is nothing but the origin that is $\left(0,0\right)$ then the equation reduces to:

  $x^2+y^2=r^2$

Completing Square method:

To apply this method on an equation of the form: $a{x}^2+bx$

Method- Make sure that the coefficient of $x^2=1$ and then, add and subtract the square of the half of coefficient of $x$ and then club the terms using formulae:

  $\begin{array}{l}{\left(a+b\right)}^2=a^2+2ab+b^2\\ {\left(a-b\right)}^2=a^2-2ab+b^2\end{array}$

Calculation:

Consider the given equation:

  $x^2+y^2+2x-6y+9=0$

It can also be written as:

  $\left(x^2+2x\right)+\left(y^2-6y\right)=-9$

Now apply Completing Square Method on the two brackets:

  $\left(x^2+2x+{\left(\frac{2}{2}\right)}^2-{\left(\frac{2}{2}\right)}^2\right)+\left(y^2-6y-{\left(\frac{-6}{2}\right)}^2+{\left(\frac{-6}{2}\right)}^2\right)=-9$

Now shift the extra terms to the right hand side

  $\left(x^2+2x+{\left(\frac{2}{2}\right)}^2\right)+\left(y^2-6y+{\left(\frac{-6}{2}\right)}^2\right)=-9+{\left(\frac{2}{2}\right)}^2+{\left(\frac{-6}{2}\right)}^2$

Simplify it to get:

  $\left(x^2+2x+1^2\right)+\left(y^2-6y+3^2\right)=-9+1+9$

Recall the identities:

  $\begin{array}{l}{\left(a+b\right)}^2=a^2+2ab+b^2\\ {\left(a-b\right)}^2=a^2-2ab+b^2\end{array}$

Use this to get:

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

Compare this with the Standard equation of circle which is

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

Where $\left(h,k\right)$ is the center and $r$ is the radius of the circle.

We get:

  $\left(h,k\right)=\left(-1,3\right)$ And $r=1$

Thus, the equation $x^2+y^2+2x-6y+9=0$ represents a circle with center $\left(-1,3\right)$ and radius $1$ .