Chapter 1.5, Problem 32AYU

To determine

To find:

1. The centre $\left(h,k\right)$ and radius of the circle $x²+y²+x+y-\frac{1}{2}=0$

2. The graph of the circle

3. Intercepts of the circle, if any

Answer to Problem 32AYU

$\text{Centre }\left(-0.5,-0.5\right);\text{ radius }=1$

$x\text{-intercepts are }\frac{\sqrt{3}-1}{2}\text{and }- \frac{\sqrt{3}+1}{2}$

$y\text{-intercepts are }\frac{\sqrt{3}-1}{2}\text{and }- \frac{\sqrt{3}+1}{2}$

Explanation of Solution

Given:

Equation of the circle $x²+y²+x+y-\frac{1}{2}=0$

Calculation:-

$x²+y²+x+y-\frac{1}{2}=0$

$(x²+x)+(y²+y)-\frac{1}{2}=0$

$\left(x²+x\right)+\left(y²+y\right)-\frac{1}{2}+\frac{1}{2}=0+\frac{1}{2}$   Add $\frac{1}{2}$ to both sides

$(x²+x+\frac{1}{4})+(y²+y+\frac{1}{4})=\frac{1}{2}+\frac{1}{2}$   Complete the square of each expression in parenthesis

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

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 $h=-\frac{1}{2},k=- \frac{1}{2}$ and $r=1$.

The circle has centre $(-\frac{1}{2},-\frac{1}{2})$ and a radius 1 units. To graph the circle, first plot the centre $(-\frac{1}{2},-\frac{1}{2})$. Since the radius is 1 units, locate four points on the circle by plotting 1 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 $x$.

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

$(x+\frac{1}{2})²+\frac{1}{4}=1$

$(x+\frac{1}{2})²+\frac{1}{4}-\frac{1}{4}=1-\frac{1}{4}$   Subtract $\frac{1}{4}$ from both sides

$(x+\frac{1}{2})²=\frac{3}{4}$

$x+\frac{1}{2}=\pm \frac{\sqrt{3}}{2}$

$\text{If }x+\frac{1}{2}=\frac{\sqrt{3}}{2}$

$x+\frac{1}{2}-\frac{1}{2}=\frac{\sqrt{3}}{2}-\frac{1}{2}$   Subtract $\frac{1}{2}$ from both sides

$x=\frac{\sqrt{3}-1}{2}$

$\text{If }x+\frac{1}{2}=-\frac{\sqrt{3}}{2}$

$x+\frac{1}{2}-\frac{1}{2}=-\frac{\sqrt{3}}{2}-\frac{1}{2}$   Subtract $\frac{1}{2}$ from both sides

$x=-\frac{\sqrt{3}-1}{2}$

$x\text{-intercepts}$ are $\frac{\sqrt{3}-1}{2}$ and $-\frac{\sqrt{3}+1}{2}$.

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

$(0+\frac{1}{2})²+(y+\frac{1}{2})²=1$

$\frac{1}{4}+(y+\frac{1}{2})²=1$

$(y+\frac{1}{2})²+\frac{1}{4}-\frac{1}{4}=1-\frac{1}{4}$   Subtract $\frac{1}{4}$ from both sides

$(y+\frac{1}{2})²=\frac{3}{4}$

$y+\frac{1}{2}=\pm \frac{\sqrt{3}}{2}$

$\text{If }y+\frac{1}{2}=\frac{\sqrt{3}}{2}$

$y+\frac{1}{2}-\frac{1}{2}=\frac{\sqrt{3}}{2}-\frac{1}{2}$   Subtract $\frac{1}{2}$ from both sides

$y=\frac{\sqrt{3}-1}{2}$

$\text{If }y+\frac{1}{2}=-\frac{\sqrt{3}}{2}$

$y+\frac{1}{2}-\frac{1}{2}=-\frac{\sqrt{3}}{2}-\frac{1}{2}$Subtract $\frac{1}{2}$ from both sides

$y=-\frac{\sqrt{3}-1}{2}$

$y\text{-intercepts}$ are $\frac{\sqrt{3} - 1}{2}$, $-\frac{\sqrt{3} + 1}{2}$.