Liceo Scientifico Isaac Newton Maths course The Circle Professor Serenella Iacino Read by Cinzia...

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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 RIFERIMENTO LEZIONE

Transcript of Liceo Scientifico Isaac Newton Maths course The Circle Professor Serenella Iacino Read by Cinzia...

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

Liceo Scientifico Isaac Newton

Maths course

The Circle

Professor

Serenella Iacino

Read by

Cinzia Cetraro

UTILIZZARE SPAZIO PER INSERIRE FOTO/IMMAGINE DI RIFERIMENTO LEZIONE

Page 2: Liceo Scientifico Isaac Newton Maths course The Circle Professor Serenella Iacino Read by Cinzia Cetraro UTILIZZARE SPAZIO PER INSERIRE FOTO/IMMAGINE DI.

P(x;y)

Cr x² + y² + a ax + b by + c c = 0

aa, b b, c 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

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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

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4) c = 0C

5) a = c = 0C

6) b = c = 0 C

x² + y² + ax + by = 0

x² + y² + by = 0

x² + y² + ax = 0

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x = 2a y =

2b

r = ( )² + ( )² - c2)

P ( xp; yp )

r

x² + y² + a ax + b by + c c = 0

1)

3) P (xp; yp)

- -

2b-

2a-

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

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P1

C

P

CP2

P

C

x² + y² + ax + by + c = 0

y = mx + k

Δ = 0Δ > 0 Δ < 0

SecantSecant TangentTangent ExternalExternal

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:

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P

RealReal

P

ImaginaryImaginary

P

CoincidentCoincident

x² + y² + a ax + b by + c c = 0

y - y = m 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:

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1) External

C C’

2) Tangent from the outside

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

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3) Secant

C C’

4) Tangent within

CC’

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

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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

CC’

CC’

CC’

SecantSecant TangentTangent ExternalExternal

radical axisradical axis

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ξ’ ξ’ : x² + y² + a’x + b’y + c’ = 0

ξ ξ : x² + y² + a x + b y + c = 0

linear combination ξ ξ + λ ξ’ ξ’ = 0

First circle

From which, substituting ξ ξ and ξ’ 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’ ) = 0λ

we can obtain through a linear combination linear combination the equation of a set of circles:set of circles:

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( 1 + λ ) ( 1 + λ ) ( 1 + λ )= 0x² + y² + x + Y+

( a + λ a’ ) ( b + λ b’ ) ( c + λ c’ )

Grouping the terms of second degree, those of first degree and the constants, and dividing by (1+ (1+ λλ ) we obtain this equation:

which, for different values of λλ, represents all the circles passing through the intersections of ξξ and ξ’ ξ’ which are the Base Points of the set

except for λ = -1λ = -1

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( a - a’ )x + ( b - b’ )y + ( c - c’ ) = 0 Radical AxisRadical Axis

In fact, for this value λ = -1λ = -1 we obtain the straight line:

Degenerate CircleDegenerate CircleCC C’C’

This Radical Axis can be considered as a particular circle having an infinite radius, and can be called Degenerate Circle

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This lesson was prepared by prof.ssa Serenella Iacino

for the Liceo Scientifico Statale “Newton” of Rome..