Graph x^2+y^2=y

$x^{2} + y^{2} = y$

Subtract $y$ from both sides of the equation.

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

Complete the square for $y^{2} - y$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = - 1$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 1}{2 \cdot 1}$

Simplify the right side.

Cancel the common factor of $- 1$ and $1$ .

Rewrite $- 1$ as $- 1 (1)$ .

$d = \frac{- 1 (1)}{2 \cdot 1}$

Cancel the common factor.

$d = \frac{- 1 \cdot 1}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 1}{2}$

Move the negative in front of the fraction.

$d = - \frac{1}{2}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 1)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$e = 0 - \frac{1}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{1}{4}$

Subtract $\frac{1}{4}$ from $0$ .

$e = - \frac{1}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - \frac{1}{2})}^{2} - \frac{1}{4}$ .

${(y - \frac{1}{2})}^{2} - \frac{1}{4}$

Substitute ${(y - \frac{1}{2})}^{2} - \frac{1}{4}$ for $y^{2} - y$ in the equation $x^{2} + y^{2} - y = 0$ .

$x^{2} + {(y - \frac{1}{2})}^{2} - \frac{1}{4} = 0$

Move $- \frac{1}{4}$ to the right side of the equation by adding $\frac{1}{4}$ to both sides.

$x^{2} + {(y - \frac{1}{2})}^{2} = 0 + \frac{1}{4}$

Add $0$ and $\frac{1}{4}$ .

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

This is the form of a circle. Use this form to determine the center and radius of the circle.

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Match the values in this circle to those of the standard form. The variable $r$ represents the radius of the circle, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from origin.

$r = \frac{1}{2}$

$h = 0$

$k = \frac{1}{2}$

The center of the circle is found at $(h, k)$ .

Center: $(0, \frac{1}{2})$

These values represent the important values for graphing and analyzing a circle.

Center: $(0, \frac{1}{2})$

Radius: $\frac{1}{2}$

Graph x^2+y^2=y

Related Questions

Support LCMGCF.com by disabling your adblocker.