Question

Write the standard form of the equation and the general form of the equation of each circle of radius $r$and center $\left(h,k\right)$.

Graph each circle.

$r=\frac{1}{2},(h,k)=\left(\frac{1}{2},0\right)$

Short answer

Standard form of the circle is ${\left(x-\frac{1}{2}\right)}^2+(y-0{)}^2={\left(\frac{1}{2}\right)}^2$

General form of the circle islocalid="1646207808827" $x^2+y^2-x=0$

Graph is as follows:

Step-by-step solution

Given information

It is given that$r=\frac{1}{2},(h,k)=\left(\frac{1}{2},0\right)$.

Step 2. Standard form of the circle

The standard form the circle with center $\left(h,k\right)$and radius $r$is$(x-h{)}^2+(y-k{)}^2=r^2$

It is given that $r=\frac{1}{2}$and $(h,k)=\left(\frac{1}{2},0\right)$.

Therefore, the standard form of the circle is

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

Step 3. General form of the circle

The general form of the circle is obtained by simplifying the standard form:

$\begin{array}{rcl}{\left(x-\frac{1}{2}\right)}^2+(y-0{)}^2& =& {\left(\frac{1}{2}\right)}^2\\ \left(x^2-2·\frac{1}{2}·x+\frac{1}{4}\right)+y^2& =& \frac{1}{4}\\ x^2+y^2-x+\frac{1}{4}& =& \frac{1}{4}\end{array}$

Subtract $\frac{1}{4}$from both sides

$x^2+y^2-x=0$

Step 4. Graph of the circle

Graph is as follows: