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Distance and Midpoint Formulas; Circles

Use the Distance Formula

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example

Try it.

Use the rectangular coordinate system to find the distance between the points \((6,4)\) and \((2,1).\)

Solution

Plot the two points. Connect the two points with a line. Draw a right triangle as if you were going to find slope.
Find the length of each leg.
Use the Pythagorean Theorem to find d, the distance between the two points.	\({a}^{2}+{b}^{2}={c}^{2}\)
Substitute in the values.	\({3}^{2}+{4}^{2}={d}^{2}\)
Simplify.	\(\ 9+16={d}^{2}\)
	\(\ 25={d}^{2}\)
Use the Square Root Property.	\(\ d=5\ d=-5\)
Since distance, d is positive, we can eliminate \(d=-5.\)	The distance between the points \((6,4)\) and \((2,1)\) is 5.

The method we used in the last example leads us to the formula to find the distance between the two points \(({x}_{1},{y}_{1})\) and \(({x}_{2},{y}_{2}).\)

When we found the length of the horizontal leg we subtracted \(6-2\) which is \({x}_{2}-{x}_{1}.\)

When we found the length of the vertical leg we subtracted \(4-1\) which is \({y}_{2}-{y}_{1}.\)

If the triangle had been in a different position, we may have subtracted \({x}_{1}-{x}_{2}\) or \({y}_{1}-{y}_{2}.\) The expressions \({x}_{2}-{x}_{1}\) and \({x}_{1}-{x}_{2}\) vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say \(|{x}_{2}-{x}_{1}|\) and \(|{y}_{2}-{y}_{1}|.\)

In the Pythagorean Theorem, we substitute the general expressions \(|{x}_{2}-{x}_{1}|\) and \(|{y}_{2}-{y}_{1}|\) rather than the numbers.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Use the Midpoint Formula

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.

Example

Try it.

Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are \((-5,-4)\) and \((7,2).\) Plot the endpoints and the midpoint on a rectangular coordinate system.

Solution

Write the Midpoint Formula.	\((\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})\)
Label the points, \((\overset{{x}_{1},{y}_{1}}{-5,-4}),(\overset{{x}_{2},{y}_{2}}{7,2})\) and substitute.	\((\frac{-5+7}{2},\frac{-4+2}{2})\)
Simplify.	\((\frac{2}{2},\frac{-2}{2})\)
	\((1,-1)\) The midpoint of the segment is the point \((1,-1).\)
Plot the endpoints and midpoint.

Both the Distance Formula and the Midpoint Formula depend on two points, \(({x}_{1},{y}_{1})\) and \(({x}_{2},{y}_{2}).\) It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, it may be easier to remember the formulas.

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, \((h,k),\) and the fixed distance is called the radius, r, of the circle.

We look at a circle in the rectangular coordinate system. The radius is the distance from the center, \((h,k),\) to a point on the circle, \((x,y).\)
To derive the equation of a circle, we can use the distance formula with the points \((h,k),\) \((x,y)\) and the distance, r.	\(\ d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}\)
Substitute the values.	\(\ r=\sqrt{{(x-h)}^{2}+{(y-k)}^{2}}\)
Square both sides.	\({r}^{2}={(x-h)}^{2}+{(y-k)}^{2}\)

This is the standard form of the equation of a circle with center, \((h,k),\) and radius, r.

Example

Try it.

Write the standard form of the equation of the circle with radius 3 and center \((0,0).\)

Solution

Use the standard form of the equation of a circle	\({(x-h)}^{2}+{(y-k)}^{2}={r}^{2}\)
Substitute in the values \(r=3,h=0,\) and \(k=0.\)	\({(x-0)}^{2}+{(y-0)}^{2}={3}^{2}\)

Simplify.	\({x}^{2}+{y}^{2}=9\)

In the last example, the center was \((0,0).\) Notice what happened to the equation. Whenever the center is \((0,0),\) the standard form becomes \({x}^{2}+{y}^{2}={r}^{2}.\)

Example

Try it.

Write the standard form of the equation of the circle with radius 2 and center \((-1,3).\)

Solution

Use the standard form of the equation of a circle.	\(\ {(x-h)}^{2}+{(y-k)}^{2}={r}^{2}\)
Substitute in the values.	\({(x-(-1))}^{2}+{(y-3)}^{2}={2}^{2}\)

Simplify.	\(\ {(x+1)}^{2}+{(y-3)}^{2}=4\)

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Graph a Circle

Any equation of the form \({(x-h)}^{2}+{(y-k)}^{2}={r}^{2}\) is the standard form of the equation of a circle with center, \((h,k),\) and radius, r. We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from x and y. In the next example, the equation has \(x+2,\) so we need to rewrite the addition as subtraction of a negative.

Example

Try it.

Find the center and radius, then graph the circle: \({(x+2)}^{2}+{(y-1)}^{2}=9.\)

Solution


Use the standard form of the equation of a circle. Identify the center, \((h,k)\) and radius, r.
	Center: \((-2,1)\) radius: 3
Graph the circle.

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of \({x}^{2},{y}^{2}\) to be one.

Example

Try it.

Find the center and radius and then graph the circle, \(4{x}^{2}+4{y}^{2}=64.\)

Solution


Divide each side by 4.
Use the standard form of the equation of a circle. Identify the center, \((h,k)\) and radius, r.
	Center: \((0,0)\) radius: 4
Graph the circle.

If we expand the equation from , \({(x+2)}^{2}+{(y-1)}^{2}=9,\) the equation of the circle looks very different.

	\({(x+2)}^{2}+{(y-1)}^{2}=9\)
Square the binomials.	\({x}^{2}+4x+4+{y}^{2}-2y+1=9\)
Arrange the terms in descending degree order, and get zero on the right	\({x}^{2}+{y}^{2}+4x-2y-4=0\)

This form of the equation is called the general form of the equation of the circle.

If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y. Then we can graph the circle using its center and radius.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Distance Formula: The distance d between the two points \(({x}_{1},{y}_{1})\) and \(({x}_{2},{y}_{2})\) is
\[d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}\]
Midpoint Formula: The midpoint of the line segment whose endpoints are the two points \(({x}_{1},{y}_{1})\) and \(({x}_{2},{y}_{2})\) is
\[(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})\]
To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.
Circle: A circle is all points in a plane that are a fixed distance from a fixed point in the plane. The given point is called the center, \((h,k),\) and the fixed distance is called the radius, r, of the circle.
Standard Form of the Equation a Circle: The standard form of the equation of a circle with center, \((h,k),\) and radius, r, is
General Form of the Equation of a Circle: The general form of the equation of a circle is
\[{x}^{2}+{y}^{2}+ax+by+c=0\]

Distance and Midpoint Formulas; Circles

Use the Distance Formula

In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try it.

\((2,0)\) and \((5,4)\)

Solution

\(d=5\)

Try it.

\((-4,-3)\) and \((2,5)\)

Try it.

\((-4,-3)\) and \((8,2)\)

Solution

Try it.

\((-7,-3)\) and \((8,5)\)

Try it.

\((-1,4)\) and \((2,0)\)

Solution

Try it.

\((-1,3)\) and \((5,-5)\)

Try it.

\((1,-4)\) and \((6,8)\)

Solution

Try it.

\((-8,-2)\) and \((7,6)\)

Try it.

\((-3,-5)\) and \((0,1)\)

Solution

\(d=3\sqrt{5},d\approx 6.7\)

Try it.

\((-1,-2)\) and \((-3,4)\)

Try it.

\((3,-1)\) and \((1,7)\)

Solution

\(d=2\sqrt{17},d\approx 8.2\)

Try it.

\((-4,-5)\) and \((7,4)\)

Use the Midpoint Formula

In the following exercises, ⓐ find the midpoint of the line segment whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.

Try it.

\((0,-5)\) and \((4,-3)\)

Solution

ⓐ Midpoint: \((2,-4)\)
ⓑ

Try it.

\((-2,-6)\) and \((6,-2)\)

Try it.

\((3,-1)\) and \((4,-2)\)

Solution

ⓐ Midpoint: \((3\frac{1}{2},-1\frac{1}{2})\)
ⓑ

Try it.

\((-3,-3)\) and \((6,-1)\)

Write the Equation of a Circle in Standard Form

In the following exercises, write the standard form of the equation of the circle with the given radius and center \((0,0).\)

Try it.

Radius: 7

Solution

\({x}^{2}+{y}^{2}=49\)

Try it.

Radius: 9

Try it.

Radius: \(\sqrt{2}\)

Solution

\({x}^{2}+{y}^{2}=2\)

Try it.

Radius: \(\sqrt{5}\)

In the following exercises, write the standard form of the equation of the circle with the given radius and center

Try it.

Radius: 1, center: \((3,5)\)

Solution

\({(x-3)}^{2}+{(y-5)}^{2}=1\)

Try it.

Radius: 10, center: \((-2,6)\)

Try it.

Radius: \(2.5,\) center: \((1.5,-3.5)\)

Solution

\({(x-1.5)}^{2}+{(y+3.5)}^{2}=6.25\)

Try it.

Radius: \(1.5,\) center: \((-5.5,-6.5)\)

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

Try it.

Center \((3,-2)\) with point \((3,6)\)

Solution

\({(x-3)}^{2}+{(y+2)}^{2}=64\)

Try it.

Center \((6,-6)\) with point \((2,-3)\)

Try it.

Center \((4,4)\) with point \((2,2)\)

Solution

\({(x-4)}^{2}+{(y-4)}^{2}=8\)

Try it.

Center \((-5,6)\) with point \((-2,3)\)

Graph a Circle

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

Try it.

\({(x+5)}^{2}+{(y+3)}^{2}=1\)

Solution

ⓐ The circle is centered at \((-5,-3)\) with a radius of 1.
ⓑ

Try it.

\({(x-2)}^{2}+{(y-3)}^{2}=9\)

Try it.

\({(x-4)}^{2}+{(y+2)}^{2}=16\)

Solution

ⓐ The circle is centered at \((4,-2)\) with a radius of 4.
ⓑ

Try it.

\({(x+2)}^{2}+{(y-5)}^{2}=4\)

Try it.

\({x}^{2}+{(y+2)}^{2}=25\)

Solution

ⓐ The circle is centered at \((0,-2)\) with a radius of 5.
ⓑ

Try it.

\({(x-1)}^{2}+{y}^{2}=36\)

Try it.

\({(x-1.5)}^{2}+{(y+2.5)}^{2}=0.25\)

Solution

ⓐ The circle is centered at \((1.5,-2.5)\) with a radius of \(0.5.\)
ⓑ

Try it.

\({(x-1)}^{2}+{(y-3)}^{2}=\frac{9}{4}\)

Try it.

\({x}^{2}+{y}^{2}=64\)

Solution

ⓐ The circle is centered at \((0,0)\) with a radius of 8.
ⓑ

Try it.

\({x}^{2}+{y}^{2}=49\)

Try it.

\(2{x}^{2}+2{y}^{2}=8\)

Solution

ⓐ The circle is centered at \((0,0)\) with a radius of 2.
ⓑ

Try it.

\(6{x}^{2}+6{y}^{2}=216\)

In the following exercises, ⓐ identify the center and radius and ⓑ graph.

Try it.

\({x}^{2}+{y}^{2}+2x+6y+9=0\)

Solution

ⓐ Center: \((-1,-3),\) radius: 1
ⓑ

Try it.

\({x}^{2}+{y}^{2}-6x-8y=0\)

Try it.

\({x}^{2}+{y}^{2}-4x+10y-7=0\)

Solution

ⓐ Center: \((2,-5),\) radius: 6
ⓑ

Try it.

\({x}^{2}+{y}^{2}+12x-14y+21=0\)

Try it.

\({x}^{2}+{y}^{2}+6y+5=0\)

Solution

ⓐ Center: \((0,-3),\) radius: 2
ⓑ

Try it.

\({x}^{2}+{y}^{2}-10y=0\)

Try it.

\({x}^{2}+{y}^{2}+4x=0\)

Solution

ⓐ Center: \((-2,0),\) radius: 2
ⓑ

Try it.

\({x}^{2}+{y}^{2}-14x+13=0\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches.
If you missed this problem, review .

Rivela la risposta

\(20\ \text{inches}\)
Factor: \({x}^{2}-18x+81.\)
If you missed this problem, review .

Rivela la risposta

\({\left(x-9\right)}^{2}\)
Solve by completing the square: \({x}^{2}-12x-12=0.\)
If you missed this problem, review .

Rivela la risposta

\(x=6\pm 4\sqrt{3}\)

Use the rectangular coordinate system to find the distance between the points \((6,4)\) and \((2,1).\)

Rivela la risposta

Plot the two points. Connect the two points with a line. Draw a right triangle as if you were going to find slope.
Find the length of each leg.
Use the Pythagorean Theorem to find d, the distance between the two points.	\({a}^{2}+{b}^{2}={c}^{2}\)
Substitute in the values.	\({3}^{2}+{4}^{2}={d}^{2}\)
Simplify.	\(\ 9+16={d}^{2}\)
	\(\ 25={d}^{2}\)
Use the Square Root Property.	\(\ d=5\ d=-5\)
Since distance, d is positive, we can eliminate \(d=-5.\)	The distance between the points \((6,4)\) and \((2,1)\) is 5.

Use the rectangular coordinate system to find the distance between the points \((6,1)\) and \((2,-2).\)

Rivela la risposta

\(d=5\)
Use the rectangular coordinate system to find the distance between the points \((5,3)\) and \((-3,-3).\)

Rivela la risposta

\(d=10\)

Use the Distance Formula to find the distance between the points \((-5,-3)\) and \((7,2).\)

Rivela la risposta

Write the Distance Formula.	\(d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}\)
Label the points, \((\overset{{x}_{1},{y}_{1}}{-5,-3}),(\overset{{x}_{2},{y}_{2}}{7,2})\) and substitute.	\(d=\sqrt{{(7-(-5))}^{2}+{(2-(-3))}^{2}}\)

Simplify.	\(d=\sqrt{{12}^{2}+{5}^{2}}\)
	\(d=\sqrt{144+25}\)
	\(d=\sqrt{169}\)
	\(d=13\)

Use the Distance Formula to find the distance between the points \((-4,-5)\) and \((5,7).\)

Rivela la risposta

\(d=15\)
Use the Distance Formula to find the distance between the points \((-2,-5)\) and \((-14,-10).\)

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\(d=13\)

Use the Distance Formula to find the distance between the points \((10,-4)\) and \((-1,5).\) Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Rivela la risposta

Write the Distance Formula.	\(d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}\)
Label the points, \((\overset{{x}_{1},{y}_{1}}{10,-4}),(\overset{{x}_{2},{y}_{2}}{-1,5})\) and substitute.	\(d=\sqrt{{(-1-10)}^{2}+{(5-(-4))}^{2}}\)
Simplify.	\(d=\sqrt{{(-11)}^{2}+{9}^{2}}\)
	\(d=\sqrt{121+81}\)
	\(d=\sqrt{202}\)
Since 202 is not a perfect square, we can leave the answer in exact form or find a decimal approximation.	\(\begin{array}{l}d=\sqrt{202} \\ \text{or} \\ d\approx 14.2\end{array}\)

Use the Distance Formula to find the distance between the points \((-4,-5)\) and \((3,4).\) Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Rivela la risposta

\(d=\sqrt{130},d\approx 11.4\)
Use the Distance Formula to find the distance between the points \((-2,-5)\) and \((-3,-4).\) Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

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\(d=\sqrt{2},d\approx 1.4\)

Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are \((-5,-4)\) and \((7,2).\) Plot the endpoints and the midpoint on a rectangular coordinate system.

Rivela la risposta

Write the Midpoint Formula.	\((\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})\)
Label the points, \((\overset{{x}_{1},{y}_{1}}{-5,-4}),(\overset{{x}_{2},{y}_{2}}{7,2})\) and substitute.	\((\frac{-5+7}{2},\frac{-4+2}{2})\)
Simplify.	\((\frac{2}{2},\frac{-2}{2})\)
	\((1,-1)\) The midpoint of the segment is the point \((1,-1).\)
Plot the endpoints and midpoint.

Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are \((-3,-5)\) and \((5,7).\) Plot the endpoints and the midpoint on a rectangular coordinate system.

Rivela la risposta
Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are \((-2,-5)\) and \((6,-1).\) Plot the endpoints and the midpoint on a rectangular coordinate system.

Rivela la risposta
Write the standard form of the equation of the circle with radius 3 and center \((0,0).\)

Rivela la risposta

Use the standard form of the equation of a circle \({(x-h)}^{2}+{(y-k)}^{2}={r}^{2}\)
Substitute in the values \(r=3,h=0,\) and \(k=0.\) \({(x-0)}^{2}+{(y-0)}^{2}={3}^{2}\)
Simplify. \({x}^{2}+{y}^{2}=9\)
Write the standard form of the equation of the circle with a radius of 6 and center \((0,0).\)

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\({x}^{2}+{y}^{2}=36\)
Write the standard form of the equation of the circle with a radius of 8 and center \((0,0).\)

Rivela la risposta

\({x}^{2}+{y}^{2}=64\)
Write the standard form of the equation of the circle with radius 2 and center \((-1,3).\)

Rivela la risposta

Use the standard form of the equation of a
circle. \(\ {(x-h)}^{2}+{(y-k)}^{2}={r}^{2}\)
Substitute in the values. \({(x-(-1))}^{2}+{(y-3)}^{2}={2}^{2}\)
Simplify. \(\ {(x+1)}^{2}+{(y-3)}^{2}=4\)
Write the standard form of the equation of the circle with a radius of 7 and center \((2,-4).\)

Rivela la risposta

\({(x-2)}^{2}+{(y+4)}^{2}=49\)
Write the standard form of the equation of the circle with a radius of 9 and center \((-3,-5).\)

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\({(x+3)}^{2}+{(y+5)}^{2}=81\)

Write the standard form of the equation of the circle with center \((2,4)\) that also contains the point \((-2,1).\)

Rivela la risposta

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center \((2,4)\) and point \((-2,1)\)

Use the Distance Formula to find the radius.	\(\ r=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}\)
Substitute the values. \((\overset{{x}_{1},{y}_{1}}{2,4}),(\overset{{x}_{2},{y}_{2}}{-2,1})\)	\(\ r=\sqrt{{(-2-2)}^{2}+{(1-4)}^{2}}\)
Simplify.	\(\ r=\sqrt{{(-4)}^{2}+{(-3)}^{2}}\)
	\(\ r=\sqrt{16+9}\)
	\(\ r=\sqrt{25}\)
	\(\ r=5\)

Now that we know the radius, \(r=5,\) and the center, \((2,4),\) we can use the standard form of the equation of a circle to find the equation.

Use the standard form of the equation of a circle.	\(\ {(x-h)}^{2}+{(y-k)}^{2}={r}^{2}\)
Substitute in the values.	\(\ {(x-2)}^{2}+{(y-4)}^{2}={5}^{2}\)
Simplify.	\(\ {(x-2)}^{2}+{(y-4)}^{2}=25\)

Write the standard form of the equation of the circle with center \((2,1)\) that also contains the point \((-2,-2).\)

Rivela la risposta

\({(x-2)}^{2}+{(y-1)}^{2}=25\)
Write the standard form of the equation of the circle with center \((7,1)\) that also contains the point \((-1,-5).\)

Rivela la risposta

\({(x-7)}^{2}+{(y-1)}^{2}=100\)
Find the center and radius, then graph the circle: \({(x+2)}^{2}+{(y-1)}^{2}=9.\)

Rivela la risposta

Use the standard form of the equation of a circle.
Identify the center, \((h,k)\) and radius, r.
Center: \((-2,1)\) radius: 3
Graph the circle.
ⓐ Find the center and radius, then ⓑ graph the circle: \({(x-3)}^{2}+{(y+4)}^{2}=4.\)

Rivela la risposta

ⓐ The circle is centered at \((3,-4)\) with a radius of 2.
ⓑ
ⓐ Find the center and radius, then ⓑ graph the circle: \({(x-3)}^{2}+{(y-1)}^{2}=16.\)

Rivela la risposta

ⓐ The circle is centered at \((3,1)\) with a radius of 4.
ⓑ
Find the center and radius and then graph the circle, \(4{x}^{2}+4{y}^{2}=64.\)

Rivela la risposta

Divide each side by 4.
Use the standard form of the equation of a circle.
Identify the center, \((h,k)\) and radius, r.
Center: \((0,0)\) radius: 4
Graph the circle.
ⓐ Find the center and radius, then ⓑ graph the circle: \(3{x}^{2}+3{y}^{2}=27\)

Rivela la risposta

ⓐ The circle is centered at \((0,0)\) with a radius of 3.
ⓑ
ⓐ Find the center and radius, then ⓑ graph the circle: \(5{x}^{2}+5{y}^{2}=125\)

Rivela la risposta

ⓐ The circle is centered at \((0,0)\) with a radius of 5.
ⓑ
ⓐ Find the center and radius, then ⓑ graph the circle: \({x}^{2}+{y}^{2}-4x-6y+4=0.\)

Rivela la risposta

We need to rewrite this general form into standard form in order to find the center and radius.

Group the x-terms and y-terms.
Collect the constants on the right side.
Complete the squares.
Rewrite as binomial squares.
Identify the center and radius. Center: \((2,3)\) radius: 3
Graph the circle.
ⓐ Find the center and radius, then ⓑ graph the circle: \({x}^{2}+{y}^{2}-6x-8y+9=0.\)

Rivela la risposta

ⓐ The circle is centered at \((\text{3},\text{4})\) with a radius of 4.
ⓑ
ⓐ Find the center and radius, then ⓑ graph the circle: \({x}^{2}+{y}^{2}+6x-2y+1=0.\)

Rivela la risposta

ⓐ The circle is centered at \((\text{-}\text{3},\text{1})\) with a radius of 3.
ⓑ

ⓐ Find the center and radius, then ⓑ graph the circle: \({x}^{2}+{y}^{2}+8y=0.\)

Rivela la risposta

We need to rewrite this general form into standard form in order to find the center and radius.


Group the x-terms and y-terms.
There are no constants to collect on the right side.
Complete the square for \({y}^{2}+8y.\)
Rewrite as binomial squares.
Identify the center and radius.	Center: \((0,-4)\) radius: 4
Graph the circle.

ⓐ Find the center and radius, then ⓑ graph the circle: \({x}^{2}+{y}^{2}-2x-3=0.\)

Rivela la risposta

ⓐ The circle is centered at \((1,0)\) with a radius of 2.
ⓑ
ⓐ Find the center and radius, then ⓑ graph the circle: \({x}^{2}+{y}^{2}-12y+11=0.\)

Rivela la risposta

ⓐ The circle is centered at \((0,6)\) with a radius of 5.
ⓑ
\((2,0)\) and \((5,4)\)

Rivela la risposta

\(d=5\)
\((-4,-3)\) and \((2,5)\)
\((-4,-3)\) and \((8,2)\)

Rivela la risposta

13
\((-7,-3)\) and \((8,5)\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Distance and Midpoint Formulas; Circles

Use the Distance Formula
Use the Midpoint Formula
Write the equation of a circle in standard form
Graph a circle

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Prova il tuo

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Più in Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value


Group the x-terms and y-terms. Collect the constants on the right side.
Complete the squares.
Rewrite as binomial squares.
Identify the center and radius.	Center: \((2,3)\) radius: 3
Graph the circle.