Topics
10. Connecting Algebra & Geometry with Coordinates
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United States of AmericaFL
Grade 912

10.06 Equations of circles

Lesson
Practice
Lesson

Concept summary

All points on a circle are the same distance from the center. The radius tells us the distance from the center to any point on the circle.

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y

We can draw a right triangle with a radius as the hypotenuse to see where the standard form for the equation of a circle comes from.

\displaystyle \left(x-h\right)^2+\left(y-k\right)^2=r^2
\bm{r}
is the radius of the circle
\bm{\left(h,k\right)}
is the center of the circle
\bm{\left(x,y\right)}
represents the coordinates of any point on the circle
Standard form of a circle

An equation that clearly shows the coordinates of the center and the radius of a circle, written as:

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

Expanded form of a circle

An equation for a circle that has been expanded, written as:

x^2+y^2+ax+by+c=0

To check whether some point \left( x_1,y_1 \right) is inside, on or outside a circle, we can compare the distance between that point and the center to the value of the radius.

Using the Pythagorean theorem, we can write these conditions as:

  • If \left(x_1-h\right)^2+\left(y_1-k\right)^2<r^2 then \left( x_1,y_1 \right) is inside the circle
  • If \left(x_1-h\right)^2+\left(y_1-k\right)^2=r^2 then \left( x_1,y_1 \right) is on the circle
  • If \left(x_1-h\right)^2+\left(y_1-k\right)^2>r^2 then \left( x_1,y_1 \right) is outside the circle

Notice that these conditions are the same as substituting the point into the equation of the circle and comparing the values on each side.

Remember that a tangent to a circle intersects the circle at exactly one point, and is perpendicular to the radius drawn from the point of tangency. So if we know the coordinates of the point of tangency and the center of the circle, we can find the gradient of a tangent line by calculating the opposite reciprocal of the gradient of the radius between those points.

Worked examples

Example 1

Consider the given circle:

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a

Determine the center of the circle.

Approach

The center of the circle is the point that is equidistant from all points on the circle.

Solution

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We know that a diameter passes through the center, so we can visualize a diameter and identify the point which is in the middle.

The center is \left(-1,2\right).

Reflection

We can also see from the sketch that the circle has a domain of -5\leq x \leq 3 and a range of -2\leq y \leq 6. The center will have coordinates that are the midpoints of the domain and range, which we can calculate to be \left( \frac{-5+3}{2},\frac{-2+6}{2} \right) = \left(-1,2\right).

b

Determine the radius, r, of the circle.

Approach

The radius of a circle is a line segment whose endpoints are the center of the circle and a point on the circle.

We can look for a point on the circle that is on the corner of the gridlines.

Solution

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We can count squares on the coordinate plane to determine that the radius is r=4.

Reflection

Notice that the domain and range of the circle are both 8 units in length. This tells us that the diameter of the circle will be 8, so the radius will be 4.

c

Write the equation of the circle.

Solution

From parts (a) and (b), we know that:

  • \left(h,k\right)=\left(-1,2\right)
  • r=4

Now, we can substitute these into the standard form of a circle.

The equation of the circle will be:\left(x--1\right)^2+\left(y-2\right)^2=4^2

which can be simplified to: \left(x+1\right)^2+\left(y-2\right)^2=16

Example 2

A circle has the equation x^2+y^2-6x+4y-12=0.

a

Rewrite the equation of the circle in standard form by completing the square.

Approach

Recall that a perfect square trinomial expands and factors to give us:

\left(x+a\right)^2=x^2+2ax+a^2

Solution

\displaystyle x^2+y^2-6x+4y-12\displaystyle =\displaystyle 0Given equation
\displaystyle x^2-6x+9 +y^2+4y+4-12\displaystyle =\displaystyle 9+4Add 9 and 4 to both sides to create perfect square trinomials
\displaystyle \left(x-3\right)^2 +\left(y+2\right)^2-12\displaystyle =\displaystyle 9+4Factor perfect square trinomials
\displaystyle \left(x-3\right)^2 +\left(y+2\right)^2\displaystyle =\displaystyle 9+4+12Add 12 to both sides
\displaystyle \left(x-3\right)^2 +\left(y+2\right)^2\displaystyle =\displaystyle 25Simplify the sum
\displaystyle \left(x-3\right)^2 +\left(y+2\right)^2\displaystyle =\displaystyle 5^2Recognize that 25 is a perfect square
b

Determine whether the point \left(1,1\right) is inside, outside, or on the circle.

Approach

We want to find the distance between the point \left( 1,1 \right) and the center of the circle. We can use the standard form found in part (a) to identify the center of the circle as \left( 3,-2 \right).

Solution

\displaystyle \left(x-3\right)^2+\left(y+2\right)^2\displaystyle =\displaystyle \left(1-3\right)^2+\left(1+2\right)^2Substitute in the point
\displaystyle =\displaystyle \left(-2\right)^2+\left(3\right)^2Simplify parentheses
\displaystyle =\displaystyle 4+9Square terms
\displaystyle =\displaystyle 13Simplify
\displaystyle 25\displaystyle >\displaystyle 13Compare to r^2

Since the distance between the point \left(1,1\right) is less than the radius, the point is inside the circle.

c

Draw a graph of this circle.

Solution

Using the equation \left(x-3\right)^2 +\left(y+2\right)^2=5^2, we can identify the center and radius.

Center: \left(3,-2\right)

Radius: r=5

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We can plot the center and then use the radius to plot the points directly left, right, up, and down from it to get four points on the circle. Then we can do our best to draw a smooth curve.

If you have a compass, we can set the compass width to the radius and draw the circle as a smooth curve about the center.

Reflection

We can use the graph of the circle to check that \left( 1,1 \right) is inside the circle.

Example 3

Determine the equation of the line tangent to the circle \left(x+1\right)^2+\left(y-3\right)^2=25 at the point \left(3, 6\right).

Approach

The tangent line will have an equation in the form y=mx+b, so we want to find the values of m and b.

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We can draw the circle, the tangent point \left(3, 6\right), and the radius from the center \left(-1,3\right) to the tangent point.

The radius that joins the center of the circle to the point \left(3, 6\right) will be perpendicular to the tangent line. This means the slope of tangent line is the negative reciprocal of the slope of the radius. So, to find the slope of the tangent line we first want to find the slope of the radius.

Once we have found the slope of the tangent line, we can substitute x=3, y=6 (the coordinates from the tangent point) into the equation of the tangent line and solve for b.

Solution

The slope of the radius is given by \dfrac{\text{change in } y}{\text{change in }x}=\dfrac{3}{4}.

If two lines are perpendicular we have m_1 \cdot m_2=-1, which means we want to solve \\ \dfrac{3}{4}\cdot m_2=-1 for m_2, the slope of the tangent line.

Solving for the slope of the tangent line we find m_2=-\dfrac{4}{3}

Next, we want to find the value of b. We can substitute x=3, y=6 and m=\dfrac{-4}{3} into the equation y=mx+b and then solve for b

\displaystyle y\displaystyle =\displaystyle mx+bEquation of a straight line
\displaystyle 6\displaystyle =\displaystyle \frac{-4}{3}\left(3\right)+bSubstitute x=3, y=6, m=\dfrac{-4}{3}
\displaystyle 6\displaystyle =\displaystyle -4+bSimplify
\displaystyle 10\displaystyle =\displaystyle bAdd 4

So the equation of the line tangent to the circle \left(x+1\right)^2+\left(y-3\right)^2=25 at the point \left(3, 6\right) is: y=-\frac{4}{3}x+10

Example 4

Grayce ties a 1 meter piece of rope around a rock and swings it in a circular motion about her outstretched arm. Grayce's swinging arm is 1.2 meters above the ground. Truc is standing 0.7 meters behind Grayce.

a

Let the location of Truc's feet be the origin. Determine the equation of the circle which describes the points that the rock passes through.

Approach

The center of the circle will be Grayce's outstretched arm, since she is swinging the rock about that point.

Since Truc is standing 0.7 meters behind Grayce, the x-value of the center will be 0.7. Since Grayce's swinging arm is 1.2 meters above the ground, the y-value of the center will be 1.2.

We are also given that the radius of the circle, the length of the rope, is 1 meter.

Solution

The equation of the circle which describes the points that the rock passes through is\left(x-0.7\right)^2+\left(y-1.2\right)^2=1

b

Determine whether or not the rock will hit the ground.

Solution

Since the rope is 1 meter long and the Grayce's arm is 1.2 meters above the ground, the lowest point that the rock can reach is 0.2 meters above the ground. Therefore, the rock will not hit the ground.

Reflection

Another way to determine that the rock does not hit the ground is to consider that the ground can be represented by the equation y=0. We can then use the following calculations:

\displaystyle \left(x-0.7\right)^2+\left(y-1.2\right)^2\displaystyle =\displaystyle 1Equation of the circle
\displaystyle \left(x-0.7\right)^2+\left(0-1.2\right)^2\displaystyle =\displaystyle 1Substitute in y=0
\displaystyle \left(x-0.7\right)^2\displaystyle =\displaystyle 1-\left(0-1.2\right)^2Subtract \left(0-1.2\right)^2 from both sides
\displaystyle \left(x-0.7\right)^2\displaystyle =\displaystyle -0.44Evaluate the right-hand side of the equation

Since \left(x-0.7\right)^2 is a perfect square, it cannot be negative. Therefore, the equation of the circle has no solutions when y=0, so the rock will not hit the ground.

c

If the rock does not hit Truc, determine his greatest possible height, rounding to the nearest centimeter.

Approach

We know that Truc is standing at x=0, so we can substitute this into the equation of the circle to find at what heights the rock passes through this position.

If Truc is any taller than the lowest height that the rock passes through this position, the rock will hit him.

Solution

\displaystyle \left(x-0.7\right)^2+\left(y-1.2\right)^2\displaystyle =\displaystyle 1Equation of the circle
\displaystyle \left(0-0.7\right)^2+\left(y-1.2\right)^2\displaystyle =\displaystyle 1Substitute in x=0
\displaystyle \left(y-1.2\right)^2\displaystyle =\displaystyle 1-\left(x-0.7\right)^2Subtract \left(0-0.7\right)^2 from both sides
\displaystyle \left(y-1.2\right)^2\displaystyle =\displaystyle 0.51Evaluate the right-hand side of the equation
\displaystyle y-1.2\displaystyle =\displaystyle \pm\sqrt{0.51}Square root both sides
\displaystyle y\displaystyle =\displaystyle \pm\sqrt{0.51}+1.2Add 1.2 to both sides
\displaystyle y=1.91,\, y\displaystyle =\displaystyle 0.49Evaluate the expression, rounding to two decimal places

Taking the lowest height that the rock passes through Truc's horizontal position, we can determine that Truc's greatest possible height, without being hit by Grayce's rock, is 0.49 meters tall.

Outcomes

MA.912.GR.3.2

Given a mathematical context, use coordinate geometry to classify or justify definitions, properties and theorems involving circles, triangles or quadrilaterals.

MA.912.GR.3.3

Use coordinate geometry to solve mathematical and real-world geometric problems involving lines, circles, triangles and quadrilaterals.

MA.912.GR.7.2

Given a mathematical or real-world context, derive and create the equation of a circle using key features.

MA.912.GR.7.3

Graph and solve mathematical and real-world problems that are modeled with an equation of a circle. Determine and interpret key features in terms of the context.

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