Chapter 1.4, Problem 29AYU

To determine

To find:

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

b. The graph of the circle

c. Intercepts of the circle, if any

Answer to Problem 29AYU

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

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

$y\text{-intercept }=-1$

Explanation of Solution

Given:

Equation of the circle $x²+y²-x+2y+1=0$

Calculation:-

$x²+y²-x+2y+1=0$

$(x²-x)+(y²+2y)+1=0$

$\left(x²-x\right)+\left(y²+2y\right)+1-1=0-1$   Subtract 1 from both sides

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

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

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

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

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

$(x-\frac{1}{2})²+1=\frac{1}{4}$

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

$(x-\frac{1}{2})²=-\frac{3}{4}$

$x-\frac{1}{2}=\pm \sqrt{\frac{-3}{4}}$   not real

No $x\text{-intercepts}$.

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

$(0-\frac{1}{2})²+\left(y+1\right)²=\frac{1}{4}$

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

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

$\left(y+1\right)²=0$

$+1=0$

$y+1-1=0–1$   Subtract 1 from both sides

$y=-1$

$y\text{-intercept }=-1$.