Chapter 1, Problem 108RE

To determine

To interpret:The given equation $2x^2+2y^2-2x+8y=\frac{1}{2}$ and decide if it represents a circle, a point or in the last case if it does not represent any graph.

Answer to Problem 108RE

The equation $2x^2+2y^2-2x+8y=\frac{1}{2}$ represents a circle with center $\left(\frac{1}{2},-2\right)$ and radius $\frac{3}{\sqrt{2}}$ .

Explanation of Solution

Given information:

The equation $2x^2+2y^2-2x+8y=\frac{1}{2}$

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:

  $2x^2+2y^2-2x+8y=\frac{1}{2}$

It can also be written as:

  $\left(2x^2-2x\right)+\left(2y^2+8y\right)=\frac{1}{2}$

Dividing throughout by $2$ to make coefficients of $x^2$ and $y^2$ as $1$ to get:

  $\left(x^2-x\right)+\left(y^2+4y\right)=\frac{1}{4}$

Now apply Completing Square Method on the two brackets:

  $\left(x^2-x+{\left(\frac{-1}{2}\right)}^2-{\left(\frac{-1}{2}\right)}^2\right)+\left(y^2+4y+{\left(\frac{4}{2}\right)}^2-{\left(\frac{4}{2}\right)}^2\right)=\frac{1}{4}$

Now shift the extra terms to the right hand side

  $\left(x^2-x+{\left(\frac{-1}{2}\right)}^2\right)+\left(y^2+4y+{\left(\frac{4}{2}\right)}^2\right)=\frac{1}{4}+{\left(\frac{-1}{2}\right)}^2+{\left(\frac{4}{2}\right)}^2$

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:

  $\begin{array}{l}{\left(x-\frac{1}{2}\right)}^2+{\left(y+\frac{4}{2}\right)}^2=\frac{1}{4}+\frac{1}{4}+4\\ {\left(x-\frac{1}{2}\right)}^2+{\left(y+2\right)}^2={\left(\frac{3}{\sqrt{2}}\right)}^2\end{array}$

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:

  $\left(h,k\right)=\left(\frac{1}{2},-2\right)$ And $r=\frac{3}{\sqrt{2}}$

Thus, equation $2x^2+2y^2-2x+8y=\frac{1}{2}$ represents a circle with center $\left(\frac{1}{2},-2\right)$ and radius $\frac{3}{\sqrt{2}}$ .