

Equations of circles and ellipses 


This article describes translation, scaling and rotation of circles and ellipses.
First some definitions:
variable  a character such as x or y, a number that may change

constant  a fixed value such as 3, 5 but also a, b or x_{1}, y_{1} 
formula  a arithmetic expression with variables 
equation  a construct with format formula = formula 
A function is a special case of an equation.
Left of the "=" character is one variable only, the result of the formula right of the "=".
The "..." stand for operations and constants.
x is called the independent and y the dependent variable.
x is assigned a value and for each value of x y may be calculated.
A function has exactly one number as result so functions may be part of a formula.
Examples of formulas are:
3x^{2} + 5y^{2}
7x  8y  10.
An example of an equation is:
3x^{2} + 5y^{2} = 2x  y + 25
Some examples of functions are:
y = 3sin(x)
y = 3x^{3}  2x^{2} + x  7
Graphics
Graphics are pictures of equations and functions.
Take equation x^{2} + y^{2} = 1
For a value of x the value of y may be calculated.
Writing these number pairs as (x,y) they become coordinates in a plane.
For each (x,y) pair we may paint a dot.
For the case above this results in:
A circle with center (0,0) and radius 1.
The Pythagoras theorem shows why.
Translation
The center of the circle may be at another place than (0,0).
What would happen to the equation in this case?
We apprpoach the translation (shifting) ploblem in a general way.
Say we have an undefined equation ..x..y.. = ..x..y.. and point (x_{1},y_{1}) is a point on the graph.
Next we replace (x) by (x+1) and (y) by (y+2).
The result is ..(x+1)..(y+2) = ..(x+1)..(y+2)
Say point (x_{2},y_{2}) is on the new graph.
This will be the case if :
(x_{2} + 1) = x_{1}
(y_{2} + 2) = y_{2}
because the equation was solved for (x_{1},y_{1})
so:
x_{2} = x_{1}  1
y_{2} = y_{1}  2
In a picture:
The graph is shifted 1 left and 2 down.
In general, if in any equation :
x is replaced by (x+h) the graph will shift h positions left.
y is replaced by (y+v) the graph will shift v positions down.
Note : For negative values of h and v the graph will shift right and up.
This is true for any function or equation.
Translation and equation
We start with circle x^{2}+y^{2}=1 and replace (x) by (x+h) and (y) by (y+v).
The result:
(x+h)^{2} + (y+v)^{2}=1
x^{2}+2hx+h^{2}+y^{2}+2vy+v^{2}=1
The factors of the squares of a and b do not change.
New are terms with x and y showing up as 2hx and 2vy.
This indicates a translation.
Example
x^{2}10x+y^{2}+6y=47 is a shifted circle.
2h = 10.....h = 5.....shift of 5 to the right.
2v = 6.......v = 3......shift of 3 down.
Scaling x,y
In equation x^{2}+y^{2} = 1 we replace x by x/2.
What will happen?
Say (x_{1},y_{1}) is on the circle and after the change this point has become (x_{2},y_{2}).
This is true if:
x_{2}/2 = x_{1}
or
x_{2} = 2x_{1}
The result is doubling of the width.
Such a stretched (oval) circle is called an ellipse.
In general, if in any equation:
x is replaced by (x/a) the graph will stretch by a factor a in horizontal direction.
y is replaced by (y/b) the graph will stretch by a factor b in vertical direction.
Note: for factors < 1 the dimensions will shrink.
The general equation of an ellipse is:
Scaling and the equation
Cicle x^{2} + y^{2} = 1 is stretched horizontally by a and vertically b resulting in (x/a)^{2} + (y/b)^{2} = 1.
Multiplication by a^{2}b^{2} makes b^{2}x^{2} + a^{2}y^{2} = a^{2}b^{2}
Scaling changes the factors of the x, squares and the constant (without x or y)
No new terms with x, y are added.
Example
Given is ellipse 9x^{2}+49y^{2} = 441.
Calculate "a" and "b".
Divide both terms by 441 :
Rotation
We rotate point P(x_{1},y_{1}) counterclockwise by angle φ
For convenience we observe x and y separately.
Q is the rotated point P.
The center of the rotation is origin (0,0).
Rotation of x_{1} causes
x_{2} = x_{1}cosφ
Δy = x_{1}sinφ.......infuence of x on y
Rotation of y_{1}:
y_{2} = y_{1}cosφ
Δx = y_{1}sinφ........influence of y on x
Adding the results:
x_{2} = x_{1}cosφ  y_{1}sinφ
y_{2} = y_{1}cosφ + x_{1}sinφ
In general:
If in any equation with variables x and y :
(x) is replaced by xcos(φ)ysin(φ)
(y) is replaced by ycos(φ+xsin(φ)
the graph will rotate φ degrees clockwise.
(Remember: the (x,y) values which originally solved the equation were located at counterclockwise rotation over degree φ)
Rotation and equation
Ellipse b^{2}x^{2}+a^{2}y^{2} = a^{2}b^{2} is rotated right by φ degrees.
The rotation center is (0,0).
The equation changes into
b^{2}(xcosφ  ysinφ)^{2} + a^{2}(ycosφ + xsinφ)^{2} = a^{2}b^{2}.
Now write this equation without parentheses and terms with xy appear
2b^{2}xysinφcosφ
2a^{2}xysinφcosφ
this is the way to recognize a rotation.
Rotation also changes the terms with x, y, x^{2}, y^{2} and the constant term.
Example
Ellipse 4x^{2} + 9y^{2} = 36 is rotated.
The new equation shows the term 2,5xy.
Question: over which angle did rotation take place?
b = 2.
a = 3.
2(a^{2}  b^{2})sinφcosφ = 2,5
10sinφcosφ = 2,5
2sinφcosφ = 0,5
sin(2φ) = 0,5
φ = 15
Example
Given is ellipse: 2x^{2}+y^{2} = 2(x + y + xy 1 ).
Below the graph is pictured:
The x,y terms are the result of translation, the xy term indicates rotation.
How much?
First we try to shift the graph such that the x and y terms disappear.
replace x by (x+h) and y by (y+v).
This results in two equations which enables solving of h and v.
2(x+h)^{2} + (y+v)^{2} = 2(x+h) + 2(x+v) +2(x+h)(y+v) 2
Extract the terms with x and y :
x terms: 4h = 2 + 2v
y terms: 2v = 2 + 2h
Simplify:
2h  v = 1
v  h = 1
Solution:
h = 2
v = 3
This is in agreement with the graph.
The new equation becomes after shifting 2 left and 3 down:
2(x+3)^{2} + (y+2)^{2} = 2[x+3 + y + 2 + (x+3)(y+2)  1]
after simplification:
2x^{2} + y^{2}  2xy = 3
Which rotation took place, which are the dimensions of the ellipse?
procedure to find the rotation
We depart from the general equation of an ellipse  

However, right of the "=" character another number then 1 may be found.
This means that in fact a and b need to be changed.
Before we noticed that rotation changes a and b, no constant change.
In the equation b^{2}x^{2} + a^{2}y^{2} = a^{2}b^{2} the number right of the "=" is the product of x^{2} and y^{2} factors.
Why is this important?
We see 2x^{2} + y^{2}  2xy = 3
Right of the "=" is no 1 but a 3. Later I come back at this pitfall.
Say
then rotation of a graph over φ degrees clockwise takes place if
(x) is replaced by (px  qy)
(y) is replaced by (py + qx)
Now we apply this operation to the standard ellipse equation.
This results in a new equation with different factors of x^{2}, y^{2} and xy terms.
Next we equate these values to those in 2x^{2} + y^{2}  2xy.
This results in a new set of equations with a,b,p,q.
The goal is to solve these equations to know the rotation and a,b, the ellipse dimensions.
The calculated a,b where derived from an equation with format ....= 1.
However our equation shows ....= 3.
Therefore we depart from equation b^{2}x^{2} + a^{2}y^{2} = 3a^{2}b^{2}.
(from the form without denominators)
Above we see the set of equations to be solved.
p^{2} + q^{2} = 1 we could add.
First we look after p and q.
Subtract row 2 from row 1 and combine terms with p^{2} en q^{2}:
Note: φ is the angle of rotation. z = tanφ
Now we look after a and b.
Again the 3 previous equations with p,q,a,b are the start.
Add rows 1 and 2 and place p^{2} and q^{2} outside parenthesis:
Right of the "=" comes 3a^{2}b^{2} = 2,946.
The equation of the ellise was before translation and rotattion:
what the next picture shows:
For clarity, two circles with radius a and b are included.
Exercises
We started this example with equation 2x^{2}+y^{2} = 2(x + y + xy  1)
But what about nx^{2}+y^{2} = 2(x + y + xy  1) for cases where n = 3,4,...
In a math Facebook group I found this problem:
solve the equation
5x^{2}+y^{2} = 2(x + y + xy  1)
this seems a weird question at first glance because we see a shifted and rotated ellipse.
The plot of this equation however shows only one single point.
What happened? Which point?

