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

Answers


Q01. 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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Q02. 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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Q03. 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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Q04. Circle in parametric equations

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Q05. 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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Q06. 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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Q07. 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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Q08. 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 ? Go to Begin

    Q09. 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 Go to Begin

    Q10. Study ex-central triangle
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    Q11. Quize for circle
    • 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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    Q12. Answers to quize for circle
    • 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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    Q13. Reference in MD2002 on PC computer
    From the keyword list find the program numbers of following keywords
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    Q14. 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.
    Go to Begin

    Q15. 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 ? Go to Begin

    Q16. Add new question
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