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Liceo Scientifico Isaac Newton Maths course The Circle Professor Serenella Iacino Read by Cinzia Cetraro UTILIZZARE SPAZIO PER INSERIRE FOTO/IMMAGINE DI.

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Presentation on theme: "Liceo Scientifico Isaac Newton Maths course The Circle Professor Serenella Iacino Read by Cinzia Cetraro UTILIZZARE SPAZIO PER INSERIRE FOTO/IMMAGINE DI."— Presentation transcript:

1 Liceo Scientifico Isaac Newton Maths course The Circle Professor Serenella Iacino Read by Cinzia Cetraro UTILIZZARE SPAZIO PER INSERIRE FOTO/IMMAGINE DI RIFERIMENTO LEZIONE

2 P(x;y) C r a b c x² + y² + ax + by + c = 0 a b ca, b, ca b ca, b, c x y We define a circle as the geometric locus of the points P (x,y) equidistant from a point C, which is the center

3 1) a = b = 0 C 2) b = 0 C C 3) a = 0 x² + y² + c = 0 x² + y² + ax + c = 0 x² + y² + by + c = 0 The different positions of a circle on a Cartesian Plane, according to the variation of its coefficients a, b, c

4 4) c = 0 C 5) a = c = 0 C 6) b = c = 0 C x² + y² + ax + by = 0 x² + y² + by = 0 x² + y² + ax = 0

5 x = 2 a y = y = 2 b r = ( )² + ( )² - c 2) P ( xp; yp ) r a bc x² + y² + ax + by + c = 0 1) 3) P (xp; yp) - - 2 b - 2 a - x y To find the equation of a circle we need three independent conditions to determine the values of the constants, that is one condition for each constant

6 P1 C P C P2 P C x² + y² + ax + by + c = 0 y = mx + k Δ = 0 Δ > 0 Δ < 0 SecantTangentExternal We can consider the position of a straight line in relation to a circle. It could be: From an algebraic point of view we must solve this system:

7 PReal P Imaginary P Coincident a b c x² + y² + ax + by + c = 0 m y - y = m ( x - x ) P P Now, we want to determine the equations of the two tangents to a circle from a given point P (xp ; yp ) From an algebraic point of view we must solve this system:

8 1)External C C’ 2)Tangent from the outside C C’C’ Let’s consider now the position of two circles relative to each other which depending on the distance between their centres can be: if the distance between their centres is greater than sum of the radii if the distance between their centres is equal to the sum of the radii

9 3)Secant C C’ 4)Tangent within C C’C’ 5)Inside C C’ if the distance between their centres is less than the sum of the radii if the distance between their centres is equal to the difference between the radii if the distance between their centres is less than the difference between the radii

10 This is the system composed by the equations of the two circles from which we obtain the equation of the radical axis with the elimination method: x² + y² + a x + b y + c = 0 x² + y² + a’x + b’y + c’ = 0 x² + y² + a x + b y + c = 0 - x² - y² - a’ x - b’ y - c’ = 0 (a-a’)x+(b-b’)y+(c-c’)= 0 (a-a’)x+(b-b’)y+(c-c’)= 0 C C’ C C SecantTangentExternal radical axis

11 ξ’ ξ’ : x² + y² + a’x + b’y + c’ = 0 ξ ξ : x² + y² + a x + b y + c = 0 linear combination ξ ξ’ ξ + λ ξ’ = 0 First circle ξ ξ’ From which, substituting ξ and ξ’, we obtain: Starting from these equations of two circles: Second circle x² + y² + a x + b y + c + ( x² + y² + a’x + b’y + c’ ) = 0 λ linear combination set of circles: we can obtain through a linear combination the equation of a set of circles:

12 ( 1 + λ ) = 0 x² + y² + x + Y+ ( a + λ a’ ) ( b + λ b’ ) ( c + λ c’ ) (1+ λ Grouping the terms of second degree, those of first degree and the constants, and dividing by (1+ λ ) we obtain this equation: λ ξξ’ which, for different values of λ, represents all the circles passing through the intersections of ξ and ξ’ which are the Base Points of the set λ = -1 except for λ = -1

13 ( a - a’ )x + ( b - b’ )y + ( c - c’ ) = 0 Radical Axis λ = -1 In fact, for this value λ = -1 we obtain the straight line: Degenerate Circle C C’ This Radical Axis can be considered as a particular circle having an infinite radius, and can be called Degenerate Circle

14 This lesson was prepared by prof.ssa Serenella Iacino. for the Liceo Scientifico Statale “Newton” of Rome.


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