Let C be the curve with parametric equations [ x = a + r cos(t) , y = b + r sin(t) ].
P(x,y) is on C
<=>
/ x = a + r cos(t)
\ y = b + r sin(t)
<=>
/ (x  a)/r = cos(t)
\ (y  b)/r = sin(t)
<=>
( (x  a)/r )^{2} + ( (y  b)/r )^{2} = 1
<=>
(x  a)^{2} + (y  b)^{2} = r^{2}
<=>
P(x,y) is on the circle with center (a,b) and radius r
Conclusion: The circle C with center (a,b) and radius r has parametric equations
[ x = a + r cos(t) , y = b + r sin(t) ].
With each t corresponds a point on the circle and vice versa.
P(a + r cos(t), b + r sin(t)) is a variable point on the circle C.
Example 1:
The circle C has an equation (x  3)^{2} + (y  6)^{2} = 25 and P(6,10) is on C.
Say M is the center of C. Find the points A and B, on C, such that PA=PB = 5.
Since the radius of the circle is 5, the triangles MPA and MPB are equilateral.
The angles are pi/3 radians.
The parametric equations of the circle are [ x = 3 + 5 cos(t) , y = 6 + 5 sin(t) ].
Point P has a tvalue t_{o} such that cos(t_{o})=0.6 and sin(t_{o})= 0.8.
From this we have t_{o} = 0.927295218
For the point A the tvalue is t_{o}+pi/3 and then A = A(1.036 , 10.598)
For the point B the tvalue is t_{o}pi/3 and then B = B(7.964 , 5.402)
Example 2:
We take an arc of a circle with parametric equations
[ x = cos(t) , y = 1 + sin(t) ] with t in [ pi/2 , 0]
and the curve y = arccos(x).
Calculate the intersection point of these curves.
Therefore we have to calculate the tvalue such that 1 + sin(t) = arccos(cos(t)).
Since t is in [ pi/2 , 0], arccos(cos(t)) = t.
So, we calculate the tvalue such that sin(t) + t + 1 = 0.
We can't solve the last equation algebraically. With a plot of sin(t) + t + 1,
we can find the first approximation t = 0.51 to the root.
With an
efficient iteration method we find a very good approximation in only 4 steps.
0.51
0.51090446454
0.510972899149
0.510973425307
0.510973429357
The requested intersection point is approximately ( 0.87 , 0.51)
Given : points P_{1}(x_{1},y_{1}); P_{2}(x_{2},y_{2}); P_{3}(x_{3},y_{3}); P_{4}(x_{4},y_{4})
No set of three points are on one line.
The four points are on one circle
<=>
There exist a circle
x^{2} + y^{2} + a x + b y + c = 0
such that the points are on that circle.
<=>
There are numbers a,b and c such that
/
 x_{1}^{2} + y_{1}^{2} + a x_{1} + b y_{1} + c = 0
 x_{2}^{2} + y_{2}^{2} + a x_{2} + b y_{2} + c = 0
 x_{3}^{2} + y_{3}^{2} + a x_{3} + b y_{3} + c = 0
 x_{4}^{2} + y_{4}^{2} + a x_{4} + b y_{4} + c = 0
\
<=>
There are numbers a,b and c such that
/
 a x_{1} + b y_{1} + c = (x_{1}^{2} + y_{1}^{2} )
 a x_{2} + b y_{2} + c = (x_{2}^{2} + y_{2}^{2} )
 a x_{3} + b y_{3} + c = (x_{3}^{2} + y_{3}^{2} )
 a x_{4} + b y_{4} + c = (x_{4}^{2} + y_{4}^{2} )
\
This is a system of four equations with 3 unknowns.
We know from the theory of systems of linear equations that
The coefficient matrix is
[ x_{1} y_{1} 1 ]
[ x_{2} y_{2} 1 ]
[ x_{3} y_{3} 1 ]
[ x_{4} y_{4} 1 ]
Because P1,P2,P3 are not on one line, the rank of this
matrix is three. The last equation is the side equation.
The characteristic determinant of this equation is
 x_{1} y_{1} 1 (x_{1}^{2} + y_{1}^{2} ) 
 x_{2} y_{2} 1 (x_{2}^{2} + y_{2}^{2} ) 
 x_{3} y_{3} 1 (x_{3}^{2} + y_{3}^{2} ) 
 x_{4} y_{4} 1 (x_{4}^{2} + y_{4}^{2} ) 
This system has a solution for a,b and c if and only if
this determinant is 0. From properties of determinants
this condition is
(x_{1}^{2} + y_{1}^{2} ) x_{1} y_{1} 1 
(x_{2}^{2} + y_{2}^{2} ) x_{2} y_{2} 1 
(x_{3}^{2} + y_{3}^{2} ) x_{3} y_{3} 1  = 0
(x_{4}^{2} + y_{4}^{2} ) x_{4} y_{4} 1 
The four points P_{1}(x_{1},y_{1}); P_{2}(x_{2},y_{2}); P_{3}(x_{3},y_{3}); P_{4}(x_{4},y_{4})
are on one circle
<=>
(x_{1}^{2} + y_{1}^{2} ) x_{1} y_{1} 1 
(x_{2}^{2} + y_{2}^{2} ) x_{2} y_{2} 1 
(x_{3}^{2} + y_{3}^{2} ) x_{3} y_{3} 1  = 0
(x_{4}^{2} + y_{4}^{2} ) x_{4} y_{4} 1 
Corollary :
Given: P1,P2,P3 are not on one line
Point P(x,y) is on the circle defined by P1,P2,P3
<=>
P,P1,P2,P3 are on a circle
<=>
(x^{2} + y^{2} ) x y 1 
(x_{1}^{2} + y_{1}^{2} ) x_{1} y_{1} 1 
(x_{2}^{2} + y_{2}^{2} ) x_{2} y_{2} 1 
(x_{3}^{2} + y_{3}^{2} ) x_{3} y_{3} 1  = 0
The equation of a circle through 3 given points
P_{1}(x_{1},y_{1}); P_{2}(x_{2},y_{2}); P_{3}(x_{3},y_{3}) is
(x^{2} + y^{2} ) x y 1 
(x_{1}^{2} + y_{1}^{2} ) x_{1} y_{1} 1 
(x_{2}^{2} + y_{2}^{2} ) x_{2} y_{2} 1 
(x_{3}^{2} + y_{3}^{2} ) x_{3} y_{3} 1  = 0
Take a point P and a circle C with center M and radius r.
The distance P,M = d.
A variable line through P intersects the circle in points A and A'.
Say MN is the perpendicular bisector of [A,A'].
Then we have (vectors in bold face)
PA . PA' = (PN + NA)(PN + NA')
= (PN + NA)(PN  NA)
2 2
= PN  NA
= P,N^{2}  N,A^{2}
Now P,N^{2} = d^{2}  M,N^{2} and N,A^{2} = r^{2}  M,N^{2}
Thus PA . PA' = d^{2}  r^{2}
This result depends only on the distance d and the radius r.
It is called the power of P relative to C.
This power is strictly positive if P is outside C, it is 0 for P on C,
and it is strictly negative for P inside C.
The power of point P(x_{o},y_{o}) relative to circle C with equation
(x  a)^{2} + (y  b)^{2}  r^{2} = 0
is
(x_{o}  a)^{2} + (y_{o}  b)^{2}  r^{2}
Example:
C : (x  2)^{2} + (y  3)^{2} = 25 and P(3,1)
The power of P relative to C is 1 + 4  25 =  20
We look for the set of all points P(x,y) such that
The power of P relative to C1 = the power of P relative to C2.
<=>
(x  a)^{2} + (y  b)^{2}  r^{2} = (x  c)^{2} + (y  d)^{2}  r'^{2}
<=>
(a  c) x + (b  d) y + k = 0
From this, we see that the set is a line.
This line is called the radical axis of the circles.
The slope of that line is (ac)/(db)
The central line connecting the centers has slope (db)/(ca).
Hence, the radical axis is orthogonal with the central line of the circles.
Example:
C1: x^{2} + y^{2} = 25
C2: (x  2)^{2} + (y  3)^{2} = 9
The radical axis is the line 4 x + 6 y  29 = 0
Take three circles C1, C2, C3.
If the radical axis of C1 and C2 meets the the radical axis of C2 and C3
in point S, then S has the same power relative to the three circles.
The point S is called the radical center of C1, C2 and C3.
Then, of course, S is on the third radical axis too.
The given solution is not 'the' solution.
Most exercises can be solved in different ways.
It is strongly recommended that you at least try to solve the
problem before you read the hidden solution.
A circle C has an equation (x3)^{2} +(y7)^{2} = 49
Find the tangent lines through point A(15,5) to C
First method:
We start with a line through A with a variable slope m.
y  5 = m (x  15)
The line is a tangent to C if and only if the two intersection points coincide.
These intersection points are the solutions of the system
/ y  5 = m (x  15)

\ (x3)^{2} +(y7)^{2} = 49
We substitute y from the first equation into the second equation.
After working out we find
(1 + m^{2}) x^{2}  (4 m + 30 m^{2} + 6) x + 225 m^{2} + 60 m  36
The two intersection points coincide if and only if the discriminant is zero.
380 m^{2}  192 m + 180 = 0
The two mvalues are 0.986 and 0.481
The tangent lines are
y  5 = 0.986 (x  15) and y  5 = 0.481 (x  15)
Second method:
The center of the circle is M(3,7).
We start with a line through A with a variable slope m.
y  5 = m (x  15)
The line is a tangent to C if and only if the distance from M to the
line is equal to the radius of the circle .
 m 3  7 + 5 15 m 
 = 7
sqrt( m^{2} + 1)
<=>
(12 m + 2)^{2} = 49 (m^{2} + 1)
This is a quadratic equation in m. We find exactly the same mvalues as above.
A variable circle C has equation
x^{2} + y^{2}  2 (t^{2}  3 t + 1) x  2 (t^{2} + 2 t) y + t = 0
The number t is a parameter.
Calculate the point P with a constant power relative to C.
How much is that power.
Say P has coordinates (r,s), then the power of P relative to C is
r^{2} + s^{2}  2 (t^{2}  3 t + 1) r  2 (t^{2} + 2 t) s + t
<=> (2 s + 2 r) t^{2} + (6 r  4 s + 1) t + r^{2} + s^{2}  2 r
This power is independent of the parameter t if and only if
2 s + 2 r = 0 and 6 r  4 s + 1= 0
<=> r = 1/10 and s = 1/10
The point P(0.1; 0.1) has a constant power relative to the
variable circle. This power is 0.22 .
The red astroid has parametric equations [ x = cos^{3}(t) , y = sin^{3}(t) ].
Find the equation of the blue circle.
If t varies point P(cos^{3}(t), sin^{3}(t)) describes the astroid.
For t = 0 we find P(1,0)
For t = pi/2 we find P(0,1)
For t = pi/4 we find P(2^{3/2},2^{3/2})
So, the radius of the circle is 1/2.
The equation of the circle is x^{2} + y^{2} = 1/4
Find the equation of the two circles with radius 5 and tangent to the parabola
y=x^{2} in point T(1,1).
In point T, the slope of the tangent line to the parabola is two. The slope
of the normal line is 1/2.
The equation of the normal line through T is y = (1/2) x + 3/2.
P(t, (1/2) t + 3/2 ) is a variable point on this normal line. t is a parameter.
P is the center of a circle if and only if
PT^{2} = 25
<=>
(t1)^{2} + ((1/2) t + 3/2 )^{2} = 25
<=>
t = 1 ± 2.sqrt(5)
Then the centers are :
P_{1} ( 1 + 2 sqrt(5) , 1  sqrt(5) )
P_{2} ( 1  2 sqrt(5) , 1 + sqrt(5) )
The circles have P_{1} and P_{2} as center and 5 as radius.
Note the value of using a parameter.
The circle C_{1} has an equation x^{2} + y^{2} = 1
The circle C_{2} has an equation (x5)^{2} + y^{2} = 9
Find the equations of the common outer tangent lines
OA = 5 ; AB = 2 ; OB = sqrt(21)
tan(t) = 2/sqrt(21) = slope OB = slope of tangent line
The equation of OB is
y = ( 2/sqrt(21) ) x
<=>
2 x  sqrt(21) y = 0
The equation of the common tangent line has the form
2 x  sqrt(21) y + k = 0 with k > 0
We calculate the kvalue such that the distance from O to the tangent line is 1
k
 = 1 <=> k = 5
sqrt( 4 + 21 )
The tangent line is 2 x  sqrt(21) y + 5 = 0
Now, find the other outer tangent line
The circle C_{1} has center M_{1}(1,1), is tangent to the xaxis and to the yaxis.
Find the equation of the circle C_{2} such that
the radius of C_{2} is > 1
the center M_{2} of C2 is on the line y=x
C_{2} is tangent to the xaxis
C_{2} is tangent to the yaxis
C_{2} is tangent to C_{1} in punt T
Make a clear drawing of the position of the circles.
Let r = the radius of C_{2}
M_{2}(r,r) => O M_{2} = r sqrt(2) (*)
O M_{1} = sqrt(2)
M_{1} T = 1
T M_{2} = r
Addition gives sqrt(2) + 1 + r = O M_{2} (**)
from (*) and (**) it follows that
r sqrt(2) = sqrt(2) + 1 + r
<=>
r (sqrt(2)1) = sqrt(2) + 1
<=>
r = 3 + 2 sqrt(2)
and the equation is
(xr)^{2} +(yr)^{2} = r^{2}
Find the equation of the circle inscribed in the triangle ABC such that :
AB is y = 0
AC is y = sqrt(3) x
BC is y =  sqrt(3) x + 10
Make an accurate figure.
We'll see that the solution is easy when there is a good cooperation between
geometry and analysis.
The triangle is isosceles with top angle C and A = A(0,0) and B = B(10/sqrt(3) , 0).
The center of the inscribed circle is on the line x = 5/sqrt(3).
The angle A is 60 degrees.
The bisector through A makes with the xaxis an angle of 30 degrees.
This bisector has equation y = (1/sqrt(3)) x.
The center M of the circle is M( 5/sqrt(3) , 5/3 ).
The radius is 5/3.
The equation is (x  5/sqrt(3))^{2} + (y  5/3)^{2} = 25/9
Take the triangle ABC with A(1,1) B(2,5) and C(7,1).
Calculate the equation of the circumscribed circle.
From the theory it follows immediately that the equation of the circumscribed circle is
In point B(0,1), the line d with equation 2x+ y = 1 is tangent to a circle
and point A(5,2) is on that circle.
Find the equation of the circle.
Draw a figure. The requested circle must satisfy two conditions:
The line d is tangent in point B
Point A is on the circle
We start with the determination of all circles that meet the first condition.
To this end we look for a geometric feature that allows us to find the center
and the radius in a simple manner.
The center M of the circle is on the perpendicular l through B to d.
This perpendicular is the line y = (1/2)x + 1.
We let M slide on the line l.
M(2t, t+ 1) is a variable point on l. t is a parameter.
The radius of the circle must be M,B.
Now, M,B^{2} = 4t^{2} + t^{2} = 5t^{2}.
The equation of the circle is of the form
(x  2 t)^{2} + (y  t  1)^{2} = 5 t^{2} (*)
The first condition is met.
Now we determine the value of t so that the second condition is met.
A(5,2) ligt op de cirkel
<=>
(5  2 t)^{2} + (2  t  1)^{2} = 5 t^{2}
<=>
...
<=>
t = 13/11
We bring this value in (*) and we have the requested circle.
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