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12. Circles
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Geometry

12.03 Circles in the coordinate plane

Lesson
Practice

Circles in the coordinate plane

We can work with circles in the coordinate plane.

Circle

A locus of points equidistant from a given point, the center.

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A circle is comprised of an infinite number of points that are equidistant from the center (h,\,k). This means all points on the circle are the same distance from the center.

The distance from the center to any of these points is the length of the radius, r.

The distance between two points on the circle, represented by a straight line passing through the center, is the length of the diameter d.

The Pythagorean theorem, or distance formula, can help us find distances in the coordinate plane.

\displaystyle d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}
\bm{d}
distance between the two points
\bm{\left(x_{1},\,y_{1}\right)}
coordinates of the first point
\bm{\left(x_{2},\,y_{2}\right)}
coordinates of the second point

The center is the midpoint of the diameter so we can use the midpoint formula to find the center when we know the endpoints of the diameter.

\displaystyle \left(x_M,\,y_M\right)=\left(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2} \right)
\bm{\left(x_M,\,y_M\right)}
coordinates of the midpoint
\bm{\left(x_1,\,y_1\right)}
coordinates of the first endpoint
\bm{\left(x_2,\,y_2\right)}
coordinates of the second endpoint

Given center (h,\,k), we can determine whether points in the coordinate plane lie on a circle.

  • 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

Examples

Example 1

The given circle has a diameter with endpoints (-1,\, 3.65) and (5,\, -1.65):

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a

Find the center of the circle.

Worked Solution
Create a strategy

A diameter has endpoints on the circle and passes through center. We know that the given line segment is a diameter, which means it passes through the center.

Notice that the points (-2,1), (2,5), (6,1) and \left(2,\,-3\right) also lie on the circle. The vertical and horizontal lines connecting these points appear to be diameters, which means they too will pass throught the center. To find the center, we can find the point where all 3 lines intersect.

Apply the idea

The point of intersection of the three diameters, and the center of the circle, is (2,\,1).

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Reflect and check

We could also use the midpoint formula to calculate the center.

\displaystyle (x_{m},\, y_{m})\displaystyle =\displaystyle \left(\dfrac{x_{1} + x_{2}}{2},\, \dfrac{y_{1} + y_{2}}{2}\right)Start with the midpoint formula
\displaystyle =\displaystyle \left(\dfrac{-1 + 5}{2},\, \dfrac{3.65 + (-1.65)}{2}\right)Substitute values from the points
\displaystyle =\displaystyle \left(\dfrac{4}{2},\, \dfrac{2}{2}\right)Evaluate each numerator
\displaystyle =\displaystyle (2,\, 1)Simplify
b

Find the length of the diameter of the circle.

Worked Solution
Create a strategy

We can find the length of the diameter by finding the distance between any two points that lie on the circle, such that the line between the two points passes through the center.

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The most efficient method for finding the length of the diameter of this circle is to find the distance of either the vertical or the horizontal diameter.

Apply the idea

Because the vertical and horizontal distances are both whole numbers, we can use either one to determine the length of the diameter. By counting the distances, we find that the diameter is 8 units long.

Reflect and check

We could also use the distance formula using the given points to calculate the diameter. Let (x_{1},\, y_{1}) be (-1,3.65) and let (x_{2},\, y_{2}) be (5,\,-1.65).

\displaystyle d\displaystyle =\displaystyle \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}Start with distance formula
\displaystyle =\displaystyle \sqrt{\left(5-(-1)\right)^{2}+\left(-1.65-3.65\right)^{2}}Substitute values from the points
\displaystyle =\displaystyle \sqrt{(6)^{2}+(-5.3)^{2}}Evaluate the subtraction
\displaystyle =\displaystyle \sqrt{36+28.09}Evaluate the exponents
\displaystyle =\displaystyle \sqrt{64.09}Evaluate the addition
\displaystyle \approx\displaystyle 8Simplify

Because the points on the circle are rounded values, there is a slight difference between the exact length of the diameter and the length we find when using the distance formula.

Example 2

Find the center of the circle whose endpoints of a diameter are \left(-1.5,\, 4\right) and \left(4.5,\,-2\right).

Worked Solution
Create a strategy

The diameter of the circle passes through the center and is twice the radius. From this, we also know that the center is the midpoint of the diameter.

Apply the idea

Recall the midpoint formula: \left(\dfrac{x_{1}+x_{2}}{2},\,\dfrac{y_{1}+y_{2}}{2}\right) Substituting the values from the endpoints of the diameter, we find the center of the circle to be:

\displaystyle \left(\dfrac{-1.5+4.5}{2},\,\dfrac{4+-2}{2}\right)\displaystyle =\displaystyle \left(1.5,\,1\right)
Reflect and check

We can plot the given endpoints and use technology to graph the circle. We can see that the center appears to be at \left(1.5,\,1\right), which confirms our answer.

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Example 3

Determine if the point (4,\,3) lies on the circle with a center at (1,\,1) and radius of 3.

Worked Solution
Create a strategy

We want to find the distance between the point \left( 4,\,3 \right) and the center of the circle and compare it to the radius. We can do this by finding the distance between the point and the center and seeing if it is equal to the radius.

Apply the idea
\displaystyle r\displaystyle =\displaystyle \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}Start with the distance formula
\displaystyle =\displaystyle \sqrt{\left(4-1\right)^{2}+\left(3-1\right)^{2}}Substitute values from the points
\displaystyle =\displaystyle \sqrt{\left(3\right)^{2}+\left(2\right)^{2}}Evaluate the subtraction
\displaystyle =\displaystyle \sqrt{9+4}Evaluate the exponents
\displaystyle =\displaystyle \sqrt{13}Evaluate the addition

Because the distance between the center and the point is not equal to the length of the radius, this point does not lie on the circle.

Reflect and check

The distance between the point \left(1,\,1\right) and \left(4,\,3\right) is approximately 3.6 units. Since this is greater than 3, the point lies outside the circle.

We can use the radius to find other points on the circle, such as:

  • The point directly right of the center: \left(1+3,\,1\right)=\left(4,\,1\right)
  • The point directly left of the center: \left(1-3,\,1\right)=\left(-2,\,1\right)
  • The point directly above the center: \left(1,\,1+3\right)=\left(1,\,4\right)
  • The point directly below the center: \left(1,\,1-3\right)=\left(1,\,-2\right)

We can use technology to graph the circle and check that these points lie on the circle, and the given point lies outside the circle.

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Example 4

A circle has a center of (-1,\,-4) and contains the point (4,\,8)

a

Find the length of the radius.

Worked Solution
Create a strategy

The distance from the center to any point on the circle is the radius. So to find the radius, we can calculate the distance from the center to the given point using the distance formula.

Apply the idea

We can use the coordinates of the two points with the distance formula: \left(x_1,\,y_1\right)=\left(-1,\,-4\right) and \left(x_2,\,y_2\right)=\left(4,\,8\right)

\displaystyle r\displaystyle =\displaystyle \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}Start with the distance formula
\displaystyle =\displaystyle \sqrt{\left(4-(-1)\right)^{2}+\left(8-(-4)\right)^{2}}Substitute values from the points
\displaystyle =\displaystyle \sqrt{\left(5\right)^{2}+\left(12\right)^{2}}Evaluate the subtraction
\displaystyle =\displaystyle \sqrt{25+144}Evaluate the exponents
\displaystyle =\displaystyle \sqrt{169}Evaluate the addition
\displaystyle =\displaystyle 13Evaluate the square root

The length of the radius is 13 units.

Reflect and check

We could have also used the Pythagoream theorem to find the length of d since the distance formula comes from the Pythagorean theorem. To do this, we can plot the points and connect them to visualize the radius.

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Then, we will find the vertical and horizontal distances between the center and the given point.

The horizontal distance is 5 and the vertical distance is 12. We can use these values in the Pythagorean theorem to find the length of the radius.

\displaystyle c^2\displaystyle =\displaystyle a^2+b^2Begin with the Pythagorean Theorem
\displaystyle c^2\displaystyle =\displaystyle 5^2+12^2Substitute vertical and horizontal lengths
\displaystyle c^2\displaystyle =\displaystyle 15+144Evaluate the exponents
\displaystyle c^2\displaystyle =\displaystyle 169Evaluate the addition
\displaystyle \sqrt{c^2}\displaystyle =\displaystyle \sqrt{169}Square root both sides
\displaystyle c\displaystyle =\displaystyle 13Evaluate

The radius is 13.

b

Find the diameter of the circle.

Worked Solution
Create a strategy

Since the radius is half of the diameter, we can double the length of the radius.

Apply the idea
\displaystyle d\displaystyle =\displaystyle 2(r)Double the radius
\displaystyle =\displaystyle 2(13)Substitute the length of the radius
\displaystyle =\displaystyle 26Evaluate the multiplication

The length of the diameter is 26

Idea summary

We can use the distance formula and midpoint formula to help determine the length and location of points on the circle and the center, radius, and diameter of a circle.

Distance formula

\displaystyle d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}
\bm{d}
distance between the two points
\bm{\left(x_{1},y_{1}\right)}
coordinates of the first point
\bm{\left(x_{2},\,y_{2}\right)}
coordinates of the second point

Midpoint formula

\displaystyle \left(x_M,\,y_M\right)=\left(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2} \right)
\bm{\left(x_M,\,y_M\right)}
coordinates of the midpoint
\bm{\left(x_1,\,y_1\right)}
coordinates of the first endpoint
\bm{\left(x_2,\,y_2\right)}
coordinates of the second endpoint

Outcomes

G.PC.4

The student will solve problems in the coordinate plane involving equations of circles.

G.PC.4bi

Solve problems in the coordinate plane involving equations of circles: i) given a graph or the equation of a circle in standard form, identify the coordinates of the center of the circle;

G.PC.4bii

Solve problems in the coordinate plane involving equations of circles: ii) given the coordinates of the endpoints of a diameter of a circle, determine the coordinates of the center of the circle.

G.PC.4biii

Solve problems in the coordinate plane involving equations of circles: iii) given a graph or the equation of a circle in standard form, identify the length of the radius or diameter of the circle.

G.PC.4biv

Solve problems in the coordinate plane involving equations of circles: iv) given the coordinates of the endpoints of the diameter of a circle, determine the length of the radius or diameter of the circle.

G.PC.4bv

Solve problems in the coordinate plane involving equations of circles: v) given the coordinates of the center and the coordinates of a point on the circle, determine the length of the radius or diameter of the circle; and

G.PC.4bvi

Solve problems in the coordinate plane involving equations of circles: vi) given the coordinates of the center and length of the radius of a circle, identify the coordinates of a point(s) on the circle.

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