Chapter 1.4, Problem 32AYU

To determine

To find:

1. The centre $\left(h,k\right)$ and radius of the circle $2x²+2y²+8x+7=0$

2. The graph of the circle

3. Intercepts of the circle, if any

Answer to Problem 32AYU

$\text{Centre }\left(-2,0\right);\text{ radius }=\frac{1}{\sqrt{2}}$

$x\text{-intercepts are }\frac{1-2\sqrt{2}}{\sqrt{2}}\text{ and} -\frac{1+2\sqrt{2}}{\sqrt{2}}$

$\text{No }y\text{-intercepts}$

Explanation of Solution

Given:

Equation of the circle $2x²+2y²+8x+7=0$

Calculation:-

$2x²+2y²+8x+7=0$

$x²+y²+4x+\frac{7}{2}=0$   Divide both sides by 2

$(x²+4x)+y²+\frac{7}{2}=0$

$(x²+4x)+y²+\frac{7}{2}-\frac{7}{2}=0-\frac{7}{2}$   Subtract $\frac{7}{2}$ from both sides

$(x²+4x+4)+y²=-\frac{7}{2}+4$   Complete the square of each expression in parenthesis

$\left(x+2\right)²+y²=\frac{-7+8}{2}$

$\left(x+2\right)²+y²=\frac{1}{2}$

$\left(x+2\right)²+y²=(\frac{1}{\sqrt{2}})²$

Compare this equation with the equation $\left(x-h\right)²+\left(y-k\right)²=r²$.

The comparison yields the information about the circle. We see that $h=-2,k=0$ and $r=\frac{1}{\sqrt{2}}$.

The circle has centre $\left(-2,0\right)$ and a radius $\frac{1}{\sqrt{2}}$ units. To graph the circle, first plot the centre $\left(-2,0\right)$. Since the radius is $\frac{1}{\sqrt{2}}$ units, locate four points on the circle by plotting $\frac{1}{\sqrt{2}}$ units to the left, to the right, up and down from the centre. These four points can be used to sketch the graph.

To find the $x\text{-intercepts}$, if any, let $y\text{= }0$ and solve for $x$.

$\left(x+2\right)²+y²=(\frac{1}{\sqrt{2}})²$

$\left(x+2\right)²+0=\frac{1}{2}$

$\left(x+2\right)=\frac{1}{2}$

$\left(x+2\right)=\pm \frac{1}{\sqrt{2}}$

$\text{If }x+2=\frac{1}{\sqrt{2}}$

$x+2-2=\frac{1}{\sqrt{2}}-2$   Subtract 2 from both sides

$x = \frac{1 - 2\sqrt{2}}{\sqrt{2}}$

$\text{If }x+2=-\frac{1}{\sqrt{2}}$

$+2-2=-\frac{1}{\sqrt{2}}-2$   Subtract 2 from both sides

$x=-(\frac{1+2\sqrt{2}}{\sqrt{2}})$

$x\text{-intercepts}$ are $\frac{1-2\sqrt{2}}{\sqrt{2}}$ and $-\frac{1+2\sqrt{2}}{\sqrt{2}}$.

To find $y\text{-intercepts}$, if any, let $x=0$ and solve for $y$.

$\left(0+2\right)²+y²=(\frac{1}{\sqrt{2}})²$

$4+y²=\frac{1}{2}$

$y²+4-4=\frac{1}{2}-4$   Subtract 4 from both sides

$y²=\frac{1-8}{2}$

$y²=\frac{-7}{2}$

$y=\pm \sqrt{\frac{-7}{2}}$   not real

No $y\text{-intercepts}$.