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Circles

Definitions and Standard Equation

General Equation of a Circle

Circle from Diameter

Circle Through Three Points

Intersection of a Line and a Circle

Intersection of Two Circles

Distance from a Point to a Circle

Coaxial Circles

Radical Axis

Parametric Equations

Explorations

The circle is the first curve we will study whose equation is of the second degree. Circles have been studied since antiquity and there exists an enormous number of interesting properties, theorems and relationships involving circles. This chapter provides the underlying analytic geometry of a circle and provides a glimpse at some of the catalog of knowledge about circles.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Definitions and Standard Equation [Top]

A circle is the locus of all points P(x,y) of the plane that have a constant distance r from a fixed point C(h,k); C is called the center and r the radius of the circle. Using the formula for the distance between two points, we find the equation of a circle in standard form to be

Two particular cases of this equation occur frequently and deserve special mention. If the center is at the origin the equation reduces to

If the x-axis contains a diameter of the circle, and the y-axis touches the circle at its extremity, then the equation becomes

Circle with center at (h,k) and radius r.

Example: Write the equation of a circle with center at (-1,1) and radius 2. Plot the circle.

Solution: The equation of the circle is . The Descarta2D representation of a circle is Circle2D[{h,k},r], where {h,k} represents the coordinates of the center point, and r is the radius of the circle.

Sketch2D[{Circle2D[{-1,1},2]},

PlotRange->{{-5,5},{-5,5}}]

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Example: Determine which of the following points are on the circle centered at (-2,1) with radius 3: (a) (3,4), (b) (1,1), (c) (-2,4).

Solution: Points whose coordinates satisfy the equation

are on the circle. The Descarta2D function IsOn2D[point,circle] will return True if the point is on the circle; otherwise, it returns False.

c1=Circle2D[{-2,1},3];

p1=Point2D[{3,4}];

p2=Point2D[{1,1}];

p3=Point2D[{-2,4}];

{IsOn2D[p1,c1],IsOn2D[p2,c1],IsOn2D[p3,c1]}

Therefore, points (b) and (c) are on the circle, and point (a) is not.

Sketch2D[{c1,p1,p2,p3}]

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Two circles are said to be concentric if their center points are coincident. Two circles are coincident if their center points are coincident and their radii are equal.

Example: Show that the two circles whose equations are and are concentric, but not coincident.

Solution: The result is obvious by inspection of the equations. The Descarta2D function IsConcentric2D[circle,circle] returns True if the two circles are concentric; otherwise, it returns False. IsCoincident2D[circle,circle] returns True if the two circles are coincident; otherwise, it returns False.

c1=Circle2D[{1,2},2]; c2=Circle2D[{1,2},3];

{IsConcentric2D[c1,c2], IsCoincident2D[c1,c2]}

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General Equation of a Circle [Top]

By expanding the standard equation of a circle with center point C(h,k) and radius r the equation may be written as

which is a special case of the general second-degree equation

where the coefficients of and are equal and there is no xy term. Therefore, a necessary and sufficient condition that represent a circle is that A=C and B=0. It is not necessary that A=C=1 since the coefficients of and being equal can be divided out, reducing them to one. The equation

is the general equation of a circle. It can be reduced to standard form by completing the squares on the - and x-terms, then on the - and y-terms yielding

or

This is the equation of a circle whose center is at (-a/2,-b/2) and whose radius is given by . The equation will be a real circle only if ; if the equation represents a single point (a circle of zero radius); and if there are no real points in the locus.

Example: Find the center and radius of the circle .

Solution: Descarta2D represents the quadratic equation

as

Quadratic2D[A,B,C,D,E,F].

The function Loci2D[quad] will convert a quadratic (second-degree) equation to a list of objects represented by the equation. The function Circle2D[quad] will return a circle directly if the quadratic is indeed a circle.

{c1=Loci2D[q1=Quadratic2D[2,0,2,-5,4,-7]],

Circle2D[q1]}

The center of the circle is (5/4,-1) and the radius is . The Descarta2D function Quadratic2D[circle] converts a circle to a quadratic equation.

Quadratic2D[ c1[[1]] ] //Simplify

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Descarta2D Hint: The Descarta2D function Simplify[quad] simplifies the coefficients of a quadratic by multiplying to remove denominators and factoring to remove common factors. The form quad //Simplify is an equivalent form of the function.

Mathematica Hint: In Mathematica the elements of a list are indicated by double square brackets surrounding the index of the element in the list. In the previous example, c1[[1]] indicates the first element in the list of objects c1.

Circle from Diameter [Top]

Consider two points and defining the end points of a diameter of a circle, C. Clearly the center of the circle, (h,k), must be the midpoint of and is given by

and the radius of the circle, r, must be one-half the distance between and :

One equation of the circle C has a particularly simple form given by

as can be verified by simplifying the equation of C in standard form.

Circle Through Three Points [Top]

Since the equation of a circle has three effective parameters (h, k, r or a, b, c), in general three conditions can be imposed upon the parameters to determine one (or more) circles. In this section we look at the case of a circle passing through three points. In a later chapter we will explore a large number of conditions for constructing circles satisfying three conditions.

We can find the equation of a circle passing through three points , and , by substituting the coordinates of the points into the standard equation for a circle yielding the three equations

= | ||

= | ||

= |

This system of equations reduces to three linear equations in three unknowns, h, k and r. Simultaneous solution of the three linear equations gives

where

and

If D=0 the points are collinear and no circle passes through the three points.

Example: Find the circle passing through the three points (1,2), (-3,1) and (0,-2).

Solution: The Descarta2D function Circle2D[point,point,point] returns a circle passing through the three points.

c1=Circle2D[p1=Point2D[{1,2}],

p2=Point2D[{-3,1}],

p3=Point2D[{0,-2}]]

Sketch2D[{p1,p2,p3,c1}]

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The quadratic equation of a circle passing through three points , and is given by the determinant equation

Example: Find the quadratic equation of the circle passing through the three points (1,2), (-3,1) and (0,-2) given in the previous example.

Solution: The Descarta2D function Quadratic2D[point,point,point] returns a quadratic representing the circle passing through the three points.

Quadratic2D[p1,p2,p3]

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Intersection of a Line and a Circle [Top]

Consider the line Ax+By+C=0 and the circle . The points of intersection of the line and circle can be determined by solving the system of these two equations in two unknowns. The coordinates of the points of intersection, and , are given by

where

If (the radius is greater than the distance from the center point to the line), then there are two distinct intersection points; if , then the two intersection points are coincident (the line is tangent to the circle); and if , then the line and the circle do not intersect.

Example: Find the two points of intersection between the line and the circle whose equations are 2x-y+3=0 and .

Solution: The Descarta2D function Points2D[line,circle] returns a list of the intersection points of the line and the circle.

pts=Points2D[l1=Line2D[2,-1,3],c1=Circle2D[{1,2},3]]

Sketch2D[{l1,c1,pts},

PlotRange->{{-3,5},{-2,6}},

CurveLength2D->15]

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Intersection of Two Circles [Top]

Two circles in a special position.

Consider two circles and . The coordinates of the two intersection points, and , of these circles can be determined by solving two equations in two unknowns. Alternately, the following geometric approach can be applied. Place the center of a circle with radius at the origin and place the center of a second circle of radius at (D,0) as shown in Figure [cir:fig17]. The equations of the two circles in standard form are clearly given by and , respectively. Solving the first equation for yields . Substituting this value of into the second equation and solving for x yields

Substituting this value for x back into the first equation and solving for yields

Let and let designate the coordinates of the intersection points in this special position. Then

If the expression under the radical in the expression for is positive, then there are two distinct intersection points; if it is zero, the two intersection points are coincident (the circles are tangent at this point); and if it is negative, the two circles do not intersect. It is easy to show algebraically that

which confirms the intuitive insight that the circles do not intersect if either the sum of the radii is greater than the distance between the centers, or the difference of the radii is less than the distance between the centers.

Two circles in arbitrary positions.

Now consider two circles in arbitrary positions with centers and as shown in Figure [cir:fig18]. The x- and y-coordinates of the intersection points can be written in terms of the distances and determined from the special position shown in Figure [cir:fig17] and are given by

where

and

Therefore, the coordinates (x,y) of the intersection points of two circles without reference to trigonometric functions are

x | = | |

y | = |

Example: Find the points of intersection between the two circles

evaluated numerically.

Solution: The Descarta2D function Points2D[circle,circle] returns a list of the intersection points of the two circles.

pts=Points2D[c1=Circle2D[{2,1},3],

c2=Circle2D[{-2,-3},4]] //N

Sketch2D[{c1,c2,pts}]

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Distance from a Point to a Circle [Top]

The distance, D, from a point to the circle is given by

The inner radical represents the distance from point P the center of the circle. The validity of the formula is easily verified by considering separately whether the point is inside, outside or on the circle.

Example: Find the distance from (2,3) to the circle .

Solution: The function Distance2D[point,circle] computes the distance between a point and a circle.

Distance2D[Point2D[{2,3}],

Circle2D[{-2,-1},1]]

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Coaxial Circles [Top]

Let and be the two points of intersection of the two circles

≡ | ||

≡ |

Consider the equation . This equation represents a circle since it is of the second degree, the coefficients of and are the same, and there is no xy term. Moreover, points and are on the circle since both points satisfy the equation C. Therefore, C represents a family (or pencil) of circles through the points of intersection of the two given circles. A particular member of this family may be determined by specifying that it satisfy one other condition. Inspection of the equation reveals that C has a center (H,K) and radius R, where

H | = | |

K | = | |

= | ||

= | ||

R | = |

Example: The two circles

≡ | ||

≡ |

determine a family of circles passing through the points of intersection of and . Plot members of the family of circles for values of κ={0,±1,±2,±3±4,±5}.

Solution: The function Circle2D[circle,circle,k,Pencil2D] returns a circle parameterized by the variable k representing the pencil of circles passing through the intersection points of two circles.

Clear[k];

c1=Circle2D[{2,1},3];

c2=Circle2D[{-2,-3},4];

c12=Circle2D[c1,c2,k,Pencil2D] //Simplify

family=Map[(c12 /. k->#)&, Range[-5,5]]

Sketch2D[{c1,c2,family}]

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Radical Axis [Top]

Let and be the equations of two distinct circles as presented in the previous section. Consider the equation . Upon simplification this equation reduces to the linear equation

This line is called the radical axis of the circles and . The radical axis possesses the following properties which we state without proof.

• It is the line of the common chord if the two circles intersect in distinct real points.

• It is the common tangent line if the circles intersect in coincident points (are tangent internally or externally).

• It is a real straight line even if the circles do not intersect in real points.

• It is the locus of points from which tangents of equal length can be drawn to the two circles.

• It is perpendicular to the line of centers of the two circles.

• It does not exist (tends to infinity) as the defining circles tend to concentricity.

• The radical axes of three circles, taken in pairs, intersect in a point called the radical center.

Example: Find the radical axis of the circles and for values of h={5,6,10,11}.

Solution: The Descarta2D function Line2D[circle,circle] returns the radical axis of the two circles.

c1=Circle2D[{4,1},4];

h={5,6,10,11};

Map[(c2[#]=Circle2D[{h[[#]],1},2])&,Range[1,4]]

Map[(radaxis[#]=Line2D[c1,c2[#]])&, Range[1,4]]

Map[Sketch2D[{c1,c2[#],radaxis[#]},

PlotRange->{{-1,12},{-4,6}}]&,

Range[1,4]]

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Mathematica Hint: The Mathematica function Range[1,4] returns the list {1,2,3,4}.

Parametric Equations [Top]

A circle may be parameterized in terms of the angle, θ, that a ray from the center to the point at the parameter value makes with the +x-axis. The resulting equations are

x=h+rcosθ and y=k+rsinθ

where (h,k) is the center of the circle and r is the radius of the circle. Values in the range 0≤θ<2π generate a complete locus of points on the circle.

Example: Generate 12 equally spaced points on the circle using the parametric equations.

Solution: The Descarta2D function Circle2D[{h,k},r][t] returns the coordinates of a point at parameter t on the circle.

c1=Circle2D[{0,0},2];

pts=Map[Point2D[c1[#]]&,2*Pi*Range[0,12]/12];

Sketch2D[{c1,pts}]

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Alternately, consider the triangle T shown in Figure [cir:fig14]. Triangle T is obviously a right triangle since

Therefore, the rational forms of the trigonometric functions for angle θ are

sinθ | = | |

cosθ | = |

Substituting these expressions into the parameterization of a circle previously given yields

These equations are called the rational parameterization of the circle and have the advantage that they can be evaluated without using trigonometric functions. Parameter values in the range 0≤t≤1 produce coordinates of points on the circle in the first quadrant, 1≤t<∞ the second quadrant, -∞<t≤-1 the third quadrant, and -1≤t≤0 in the fourth quadrant. The point at θ=π radians cannot be generated using these equations, so they are generally applied only to coordinates in the first quadrant. Also, notice that the points generated by these parametric equations do not produce equally spaced points measured by distance along the circle for equally spaced parameter values.

Rational sinθ and cosθ.

Example: Plot nine points at equal parameter values on the circle in the first quadrant using the rational parametric equations of the circle.

Solution: The points can be generated directly from the equations using parameter values 0, , , , , , , and 1 .

c1=Circle2D[{0,0},5];

pts=Map[Point2D[5{1-#^2,2#}/(#^2+1)]&,Range[0,8]/8];

Sketch2D[{c1,pts},PlotRange->{{-1,6},{-1,6}}]

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Explorations [Top]

Polar Equation of a Circle. [polarcir.html]

Show that the polar equation of a circle centered at with radius R is given by

Angle Inscribed in a Semicircle. [rtangcir.html]

Show that an angle inscribed in a semicircle is a right angle.

Chord Length of Intersecting Circles. [chdlen.html]

Show that the distance, d, between the intersection points of two circles is given by

where D is the distance between the centers of the circles, and and are the radii of the two circles.

Johnson's Congruent Circle Theorem. [johnson.html]

Take any three circles , and which pass through the origin, have equal radii, r, and intersect in pairs in two distinct points (one of the points is, by construction, the origin). Prove that the circle passing through the other three points of intersection between the circles taken in pairs is congruent to the original three circles (that is, this circle has a radius of r).

Radical Center. [radcntr.html]

Prove that the radical axes of three circles taken in pairs intersect in a common point. This point is called the radical center of the three circles.

Radical Axis of Two Circles. [radaxis.html]

Show that the two circles and have the radical axis x-y=0.

Circle-Point Midpoint Theorem. [cirptmid.html]

Show that the locus of midpoints from a fixed point to a circle of radius , is a circle of radius . Furthermore, show that the center point of the locus is the midpoint of the segment between and the center of .

Circle Through Three Points. [cir3pts.html]

Show that the equation of the circle through the three points (0,0), (a,0) and (0,b) is .

Construction of Two Related Circles. [tnlncir.html]

Prove that if OP and OQ are the tangent lines from (0,0) to the circle

then the equation of circle OPQ is

Circle of Apollonius. [apollon.html]

Show that the locus of a point P(x,y) that moves so that the ratio of its distance from two fixed points and is a circle with radius

and center

where . The locus is called the Circle of Apollonius for the points and and the ratio k.

Carlyle Circle. [carlyle.html]

Given a circle, , passing through the three points (0,1), (0,-p) and (s,-p), show that the x-coordinates of the intersection points and of with the x-axis are the roots of the quadratic equation .

Castillon's Problem. [castill.html]

Let , and be three points inside the circle . Describe a method for inscribing a triangle inside such that the sides of the triangle pass through the three given points.

Radical Axis Ratio. [raratio.html]

Show that the point of intersection of the radical axis and the line of centers of two circles of radii and divides the segment between the two centers into the ratio

where d is the distance between the centers.

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