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Mathematics Dictionary
Dr. K. G. Shih
Circle in Analytic Geometry
Subjects


  • AN 05 00 | - Outlines
  • AN 05 01 | - Definition
  • AN 05 02 | - Equation of circle
  • AN 05 03 | - Equations of circle in polar form
  • AN 05 04 | - Equation of circles in parametric equations
  • AN 05 05 | - Circle equation in implicit form
  • AN 05 06 | - Locus of circle
  • AN 05 07 | - Locus of arc
  • AN 05 08 | - Locus of in-center of triangle
  • AN 05 09 | - Locus of gravity center of triangle
  • AN 05 10 | - Study ex-central triangle
  • AN 05 11 | - Circle in gemetry
  • AN 05 12 | -
  • AN 05 13 | - Reference in MD2002 on PC computer
  • AN 05 14 | - Three points define a circle
  • AN 05 15 | - Draw a circle which will tangent the sides of a triangle
  • AN 05 16 | - New
  • AN 05 17 | - Subjects of circle on internet
  • AN 05 18 | - Quiz
  • AN 05 19 | - Answer for quiz
    • Answers


    AN 05 01. Definition

    • A point C is fixed. One point P moves so that the distance PC is constant.
    • What is the locus of P ?
    • It is a circle with center C and radius PC.
    The equation of the locus
    • Let C is at (h,k) and P is at (x,y).
    • The distance between P and C is PC = Sqr((x-h)^2+(y-k)^2))
    • Square both sides : PC^2 = (x-h)^2 + (y-k)^2.
    • This the standard form of equation of circle in rectangular form.
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    AN 05 02. Equations in rectangular coordinates

    • Eqaution : (x-h)^2 + (y-k)^2 = r^2.
    • Where (h,k) is the center and r is radius.
    Unit circle in trigonometry
    • The equation is x^2 + y^2 = 1
    • Center is at (0,0) and radius is 1.
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    AN 05 03. Equations in polar coordinates

    • R = c is the equation in polar form
    Equations in trigonometry
    • Equation : r = sin(A)
      • Center at (0,1/2).
      • Radius is 1/2.
    • Equation : r = cos(A)
      • Center at (1/2,0).
      • Radius is 1/2.
    Example 1 : Prove that r = sin(A) is a circle.
    • Since r^2 = x^2 + y^2 and y = r*sin(A).
    • Hence r = y/r.
    • Hence r^2 = y.
    • Hence x^2 + y^2 - y = 0.
    • Using completing square we have
    • x^2 + (y^2 - y + 1/4 - 1/4) = 0
    • x^2 + (y-1/2)^2 = (1/2)^2
    • Hence center at (0,1/2) and radius is 1/2.
    Example 2 : Prove that r = cos(A) is a circle.
    • Since r^2 = x^2 + y^2 and x = r*cos(A).
    • Hence r = x/r.
    • Hence r^2 = x.
    • Hence x^2 + y^2 - x = 0.
    • Using completing square we have
    • y^2 + (x^2 - x + 1/4 - 1/4) = 0
    • (x-1/2)^2 + y^2 = (1/2)^2
    • Hence center at (1/2,0) and radius is 1/2.
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    AN 05 04. Circle in parametric equations

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    AN 05 05. Equation in implicit form F(x,y) = 0

    • Let equation be x^2 + y^2 + a*x + b*y + c = 0.
    • We can find the center and radius by using completing the square.
    Example 1 : Find center and radius of x^2 + y^2 + 2*x - 4*y - 4 = 0.
    • (x^2 + 2*x + 1 - 1) + (y^2 - 4*y + 4 - 4) - 4 = 0
    • (x + 1)^2 + (y - 2)^2 = 9.
    • Hence center is at (-1,2) and radius is 3.
    Example 2 : Find locus of of x^2 + y^2 + 2*x - 4*y + 5 = 0.
    • (x^2 + 2*x + 1 - 1) + (y^2 - 4*y + 4 - 4) + 5 = 0
    • (x + 1)^2 + (y - 2)^2 = 0.
    • Hence the locus is a point.
    Example 3 : Find locus of of x^2 + y^2 + 2*x - 4*y + 10 = 0.
    • (x^2 + 2*x + 1 - 1) + (y^2 - 4*y + 4 - 4) + 10 = 0
    • (x + 1)^2 + (y - 2)^2 = -5.
    • Hence the locus is not existed in real number system.
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    AN 05 06. Locus of circle

    Two points A and B are fixed.
    • 1. Point P moves with angle PAB = 90 degrees. Locus is circle.
    • 2. Point P moves with PA^2 + PB^2 = AB^2. Locus is circle.

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    AN 05 07. Locus of arc
    Two points A and B are fixed.
    • Point P moves with angle PAB = constant.
    • Locus is an arc passing the chord AB of a circle.

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    AN 05 08. Locus of in-center of triangle

    Two points A and B are fixed.
    • Point P moves with angle PAB = constant.
    • What is the locus of the incenter of triangle APB.
    The locus of the incenter is an arc.
    • Let in-center be I.
    • Angles AIB + IAB + IBA = 180
    • Angle AIB = 180 - (IAB + IBA) = 180 - (A + B)/2 = 90 + APB/2.
    • Hence A and B fixed and angle AIB fixed. The locus is a circle.
    What is in-center ?
  • Study Subjects How to draw incenter of a triangle ?
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    AN 05 09. Locus of gravity center of triangle

    Two points A and B are fixed.
    • Point P moves with angle PAB = constant.
    • What is the locus of the gravity center of triangle APB.
    The locus of the gravity center is an arc.
    • Let gravity center be G.
    • Draw DG parallel to PA. Draw EG parallel to PB.
    • Hence Angle DGE = angle APB.
    • Since AD = 2*AM/3 and BE = 2*BE/3.
    • Hence D and E are fixed and angle DGE fixed. The locus is a circle.
    What is graivty center ?
  • Study Subjects Prove median AM : AG = 3 : 2
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    AN 05 10. Study ex-central triangle
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    AN 05 11. Quize for circle

    Circles
    • In-circle of triangle ABC
      • The sides of triangle are tangents of in-circle
      • Tangent from vertex A to in-circle is (s -a)
      • Radius of in-circle is r = (s-a)*tan(A/2)
    • Circum-circle of triangle ABC
      • The Verteices triangle are on the circle
      • Radius of in-circle is r = a/(2*sin(A))
    • Centroid of triangle
      • It related with the nine point circle (GE 19 00)
    • Ortho-center of triangle
      • It related with the nine point circle (GE 19 00)
      • Two vertices and two feet of altitudes are concyclic
      • It is related with pedal triangle
    • Ex-circle of trinagle
      • Tringle has three ex-circles
      • The sides of triangle are tangents to ex-circle as well as to in-circle
      • It is also related with pedal triangle

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    AN 05 12. New


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    AN 05 13. Reference in MD2002 on PC computer

    From the keyword list find the program numbers of following keywords

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    AN 05 14. Three points define a circle

    Geometric method : Ex-center theory
     Algebra method : Solve 3 linear equations
    • Let the equation of circle be x^2 + y^2 + a*x + b*y + c = 0
    • Substitute the given points into above equation.
    • We get 3 linear equations with unknown a, b, c.
    • Solve 3 linear equation and we obtain a, b, c.
    • Then we can use completing the square to find center (h,k) and radius r.
    Example : Find equation of circle which passes (-3,4), (3,4) and (3,-4)
    • Let equation be x^2 + y^2 + a*x + b*y + c = 0
    • Substiture 3 points into this equation
      • 1. 25 - 3*a + 4*b + c = 0
      • 2. 25 + 3*a + 4*b + c = 0
      • 3. 25 + 3*a - 4*b + c = 0
    • Solve these 3 linear equation
      • Equation 2 - equation 1 we have 6*a = 0. Hence a = 0.
      • Equation 2 - equation 3 we have 8*b = 0. Hence b = 0.
      • Substiture a abd b into equation 1, we have c = -25.
      • Hence equation of circle is x^2 + y^2 = 5^2 with center at (0,0) and radius r=5.

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    AN 05 15. Draw a circle which tangent the sides of a triangle

    Draw an in-circle
    • Draw a triangle
    • Draw bisectors of three interiol angles.
    • The bisectors are concurrent at in-center I.
    • In-center to three sides havs same distance r.
    • Hence we can draw a circle with center I and radisu r.
    Draw an es-circle
    • Draw a triangle
    • Draw bisectors of one interiol angles and two exteriol angles.
    • The bisectors are concurrent at es-center E.
    • En-center to three sides havs same distance r.
    • Hence we can draw a circle with center E and radisu r.
    Study Subjects What is in-center ? What is es-center ?
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    AN 05 16. New


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    AN 05 17. Subjects of circle on internet

  • Subjects | Definition and examples of circle on internet

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    AN 05 18. Quiz

    Questions
    • 1. What is the locus of x^2 + (y-2)^2 = 3^2 ?
    • 2. What is the locus of x^2 + y^2 + 6*x - 4*y - 3 = 0.
    • 3. What is the locus of x^2 + y^2 + 6*x - 4*y + 13 = 0.
    • 4. What is the locus of x^2 + y^2 + 6*x - 4*y + 23 = 0.
    • 5. Find center and radius of x^2 + y^2 - 2*x + 4*y - 4 = 0.
    • 6. What is the curve of r = sin(A) form 0 to 180 degrees in polar coodrinates ?
    • 7. P is moving point and C(2,-3) is fixed. Find equation of locus if PC = 3.
    • 8. Describe how to draw a circle which passes three given point.
    • 9. Describe how to draw a circle which tangents three sides of triangle.
    • 10 A and B are fixed. Angle APB = constant. Find locus of ex-center if P moves.

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    AN 05 19. Answer for Quiz

    Answers
    • 1. What is the locus of x^2 + (y-2)^2 = 3^2 ?
      • It is a circle.
      • Center at (0,2) and radisu is 3.
    • 2. What is the locus of x^2 + y^2 + 6*x - 4*y - 3 = 0.
      • (x^2 + 6*x + 9 - 9) + (y^2 - 4*y + 4 - 4) - 3 = 0.
      • (x + 3)^2 + (y - 2)^2 = 4^2.
      • Hence it is a circle. Center at (-3,2) and radius = 4.
    • 3. What is the locus of x^2 + y^2 + 6*x - 4*y + 13 = 0.
      • (x^2 + 6*x + 9 - 9) + (y^2 - 4*y + 4 - 4) + 13 = 0.
      • (x + 3)^2 + (y - 2)^2 = 0.
      • Hence it is a point at (-3,2).
    • 4. What is the locus of x^2 + y^2 + 6*x - 4*y + 23 = 0.
      • (x^2 + 6*x + 9 - 9) + (y^2 - 4*y + 4 - 4) +23 = 0.
      • (x + 3)^2 + (y - 2)^2 = -10.
      • Hence it is not existed in real number system.
    • 5. Find center and radius of x^2 + y^2 - 2*x + 4*y - 4 = 0.
      • (x^2 - 2*x + 1 - 1) + (y^2 + 4*y + 4 - 4) - 4 = 0.
      • (x - 1)^2 + (y + 2)^2 = 3^2.
      • Hence it is a circle. Center at (1,-2) and radius = 3.
    • 6. What is the curve of r = sin(A) form 0 to 180 degrees in polar coodrinates ?
      • A = 0 and r = 0; A = 30 and r = 0.5; A = 60 and r = 0.866; A = 90 and r = 1.
      • A = 120 and r = 0.5; A = 150 and r = 0.5; A = 180 and r = 0.
      • In polar coordinates it is a circle with center at (0,0.5) and radius = 0.5.
    • 7. P is moving point and C(2,-3) is fixed. Find equation of locus if PC = 3.
      • Equation of locus is (x - 2)^2 + (y + 3)^2 = 3^2.
    • 8. Describe how to draw a circle which passes three given point.
      • Using the points draw a triangle ABC.
      • Draw bisectors of the sides and they meet one point which is ex-center E.
      • Ex-center has same distance to the vertices of the triangle.
      • Use ex-center as center and EA as radius to draw a circle.
    • 9. Describe how to draw a circle which tangents three sides of a triangle.
      • Draw a triangle ABC.
      • Draw bisectors of the internal angles and they meet one point.
      • The point is called in-center I which has same distance to the sides.
      • Use in-center as center and ID as radius to draw a circle
      • Where ID perpendicular to AB.
    • 10 A and B are fixed. Angle APB = constant. Find locus of ex-center if P moves.
      • Since ex-center E has same distance from A, B and P.
      • Hence EA = EB = EP when P moves. Hence E can not move (It is fixed).

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    Q00. Outline

    Equation of circle
    • Rectangular coordinater : (x-h)^ + (y-k)^2 = r^2
      • Center is at (h,k)
      • Radius is r
    • Parametric equation : x = h + r*cos(t) and y = k + r*sin(t)
      • Center is at (h,k)
      • Radius is r
      • Change to (x-h)^2 + (y-k)^2 = r^2 using cos(t)^2 + sin(t)^2 = 1
    • Implicit form : x^2 + y^2 + A*x + B*y + C = 0
      • Find center and radius
        • Use completing square
    • Special form : R = sin(A)
      • Center is at (0,1/2)
      • Radius is 1/2
    • Special form : R = cos(A)
      • Center is at (1/2,0)
      • Radius is 1/2
    Diagrams     Locus
    • Moving point (x,y) to fixed point (x0,y0) keeps same distance
      • Using distance formula, we have r = Sqr((x-x0)^2 + (y-y0)^2)
      • Square both sides
      • we have (x-x0)^2 + (y-y0)^2 = r^2
    • Moving point C and two fixed point A and B keeps angle ACB = 90 degrees
      • Triangle ABC inscribes in a circle.
      • If AB is the diameter, then angle ACB is right angle
      • This is the first mathematical law defined by 200 B.C.
    • Moving point C and two fixed point A and B keeps AC^2 + BC^2 = AB^2
      • By Pythagorean law for triangle ABC, we see that angle ACB is right angle
      • Hence locus of point C is a circle
    • Locus of x^2 + y^2 + A*x + B*y + C = 0
      • By completing the square, we can express this equation
      • (x-h)^2 + (y-k)^2 = r^2
      • It is a circle if r is GT 0
      • It is a point if if r EQ 0
      • It does not exist in real number system if r LT 0
    • Example : Find locus of x^2 + y^2 + 4*x - 2*y - 4 = 0
      • Using completing the square
      • (x^2 + 4*x + 4 - 4) + (y^2 - 2*y + 1 - 1) - 4 = 0
      • (x + 2)^2 + (y - 1)^2 = 9
      • Hence it is a circle with center at (-2,1) and radius is 3

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