Algebra Examples

$(1, 1)$ , $(1, 2)$

Step 1

The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circumference of the circle. The given end points of the diameter are $(1, 1)$ and $(1, 2)$ . The center point of the circle is the center of the diameter, which is the midpoint between $(1, 1)$ and $(1, 2)$ . In this case the midpoint is $(1, \frac{3}{2})$ .

Tap for more steps...

Step 1.1

Use the midpoint formula to find the midpoint of the line segment.

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Step 1.2

Substitute in the values for $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ .

$(\frac{1 + 1}{2}, \frac{1 + 2}{2})$

Step 1.3

Add $1$ and $1$ .

$(\frac{2}{2}, \frac{1 + 2}{2})$

Step 1.4

Divide $2$ by $2$ .

$(1, \frac{1 + 2}{2})$

Step 1.5

Add $1$ and $2$ .

$(1, \frac{3}{2})$

Step 2

Find the radius $r$ for the circle. The radius is any line segment from the center of the circle to any point on its circumference. In this case, $r$ is the distance between $(1, \frac{3}{2})$ and $(1, 1)$ .

Tap for more steps...

Step 2.1

Use the distance formula to determine the distance between the two points.

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Step 2.2

Substitute the actual values of the points into the distance formula.

$r = \sqrt{{(1 - 1)}^{2} + {(1 - \frac{3}{2})}^{2}}$

Step 2.3

Simplify.

Tap for more steps...

Step 2.3.1

Subtract $1$ from $1$ .

$r = \sqrt{0^{2} + {(1 - \frac{3}{2})}^{2}}$

Step 2.3.2

Raising $0$ to any positive power yields $0$ .

$r = \sqrt{0 + {(1 - \frac{3}{2})}^{2}}$

Step 2.3.3

Write $1$ as a fraction with a common denominator.

$r = \sqrt{0 + {(\frac{2}{2} - \frac{3}{2})}^{2}}$

Step 2.3.4

Combine the numerators over the common denominator.

$r = \sqrt{0 + {(\frac{2 - 3}{2})}^{2}}$

Step 2.3.5

Subtract $3$ from $2$ .

$r = \sqrt{0 + {(\frac{- 1}{2})}^{2}}$

Step 2.3.6

Move the negative in front of the fraction.

$r = \sqrt{0 + {(- \frac{1}{2})}^{2}}$

Step 2.3.7

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Tap for more steps...

Step 2.3.7.1

Apply the product rule to $- \frac{1}{2}$ .

$r = \sqrt{0 + {(- 1)}^{2} {(\frac{1}{2})}^{2}}$

Step 2.3.7.2

Apply the product rule to $\frac{1}{2}$ .

$r = \sqrt{0 + {(- 1)}^{2} (\frac{1^{2}}{2^{2}})}$

Step 2.3.8

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

$r = \sqrt{0 + 1 (\frac{1^{2}}{2^{2}})}$

Step 2.3.9

Multiply $\frac{1^{2}}{2^{2}}$ by $1$ .

$r = \sqrt{0 + \frac{1^{2}}{2^{2}}}$

Step 2.3.10

One to any power is one.

$r = \sqrt{0 + \frac{1}{2^{2}}}$

Step 2.3.11

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

$r = \sqrt{0 + \frac{1}{4}}$

Step 2.3.12

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

$r = \sqrt{\frac{1}{4}}$

Step 2.3.13

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$r = \frac{\sqrt{1}}{\sqrt{4}}$

Step 2.3.14

Any root of $1$ is $1$ .

$r = \frac{1}{\sqrt{4}}$

Step 2.3.15

Simplify the denominator.

Tap for more steps...

Step 2.3.15.1

Rewrite $4$ as $2^{2}$ .

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

Step 2.3.15.2

Pull terms out from under the radical, assuming positive real numbers.

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

Step 3

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ is the equation form for a circle with $r$ radius and $(h, k)$ as the center point. In this case, $r = \frac{1}{2}$ and the center point is $(1, \frac{3}{2})$ . The equation for the circle is ${(x - (1))}^{2} + {(y - (\frac{3}{2}))}^{2} = {(\frac{1}{2})}^{2}$ .

${(x - (1))}^{2} + {(y - (\frac{3}{2}))}^{2} = {(\frac{1}{2})}^{2}$

Step 4

The circle equation is ${(x - 1)}^{2} + {(y - \frac{3}{2})}^{2} = \frac{1}{4}$ .

${(x - 1)}^{2} + {(y - \frac{3}{2})}^{2} = \frac{1}{4}$

Step 5

Enter YOUR Problem