x^{2} + y^{2} + 2 m x + 2 n y + c = 0 (3)
with
m^{2} + n^{2} - c > 0
is the equation of a circle.
Proof:
We try to transform (3) to the form (1).
x^{2} + y^{2} + 2 m x + 2 n y + c + m^{2} + n^{2} = m^{2} + n^{2}
<=>
(x + m)^{2} + (y + n)^{2} = m^{2} + n^{2} - c
If m^{2} + n^{2} - c > 0 then we can write m^{2} + n^{2} - c = r^{2}
The equation (3) can be written in the form
(x + m)^{2} + (y + n)^{2} = r^{2}
So, it is the equation of the circle with center M(-m,-n)
____________
| 2 2
and radius r = \| m + n - c
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 :
A circle C with center M(3,-2) and radius 4 has parametric equations
x = 3 + 4 cos(t)
y = -2 + 4 sin(t)
Each t-value corresponds with a point of the circle.
Exercise: Draw the circle C. Choose a few t-values and calculate the coordinates
of the corresponding point P. Check if point P is on the circle C.
Example 2 :
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 t-value 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 t-value is t_{o}+pi/3 and then A = A(1.036 , 10.598)
For the point B the t-value is t_{o}-pi/3 and then B = B(7.964 , 5.402)
The coordinates of the intersection points of a circle and a line
are the solutions of the system formed by the resp. equations.
If that system has just one solution, the line is a tangent line
of the circle.
Example:
C : (x - 9)^{2} + (y - 6)^{2} = 25 (4)
a : y = -2 x + 14 (5)
(5) in (4) gives
(x - 9)^{2} + (14 - 2 x - 6)^{2} = 25
<=>
-5 x^{2} + 50 x - 120 = 0
<=>
x = 4 or x = 6
Then
y = 6 or y = 2
Take the circle C with center M(a,b) and radius r.
C: (x - a)^{2} + (y - b)^{2} = r^{2}
and point P(x_{o},y_{o}) is on C.
b - y_{o}
MP has slope ------
a - x_{o}
a - x_{o}
The tangent line has slope - ------
b - y_{o}
The tangent line has equation
a - x_{o}
y - y_{o} = - ------ (x - x_{o})
b - y_{o}
Example :
Let C be the circle with center M(3,4) and radius r = 5.
Find the equation of the tangent line in point O(0,0).
Solution:
The equation of the circle is (x-3)^{2} + (y-4)^{2} = 25
The tangent line in point O(0,0) is
3 - 0
y - 0 = - ------- (x - 0)
4 - 0
<=> y = (-3/4) x
<=> 3x + 4 y = 0
The circumference of a circle with radius r is 2.pi.r
Consider a circular arc corresponing with a central angle of t radians.
The length of the arc is directly proportional to the angle t. The length is
t
(2 pi r) . ------ = r t
2 pi
A circle has radius r.
The length of the arc, corresponing with a central angle of t radians, is r.t
The area of a circle with radius r is pi.r^{2}
Consider a circle sector with a central angle of t radians.
The area of the sector is directly proportional to the angle t.
The area is
t
(pi r^{2}) . ------ = (1/2) t r^{2}
2 pi
A circle has radius r.
The area of the sector with a central angle of t radians is (1/2) t r^{2}
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.
Find the intersection points of the parabola y = x^{2} and the circle with parametric
equations
[ x = 4 cos(t) , y = 4 + 4 sin(t) ]
First method
The circle has cartesian equation x^{2} + (y-4)^{2} = 16.
We have to find the solutions of the system
x^{2} + (y-4)^{2} = 16
y = x^{2}
The intersection points are (0,0) (sqrt(7),7) and (-sqrt(7),7)
Second method
A variable point P( 4 cos(t), 4 + 4 sin(t) ) on the circle is on the parabola
<=> 4 + 4 sin(t) = 16 cos^{2}(t)
<=> 1 + sin(t) = 4 ( 1 - sin^{2}(t))
<=> 4 sin^{2}(t) + sin(t) -3 = 0
This is a quadratic equation in sin(t)
<=> sin(t) = -1 or sin(t) = 3/4
If sin(t) = -1 then cos(t) = 0 and the intersection point is
P( 4 cos(t), 4 + 4 sin(t) ) = (0,0)
If sin(t) = 3/4 then cos(t) = ± sqrt(7)/4 and the intersection point is
P( 4 cos(t), 4 + 4 sin(t) ) = (± sqrt(7), 7)
The center M(3,4) of the circle C_{2} is on the circle C_{1} with
center O(0,0). The circle C_{2} has a radius r < 10.
The circles have common points A and B.
The points A and B are the solutions of the system
/ x^{2} + y^{2} - 6x - 8y - r^{2} + 25 = 0
\ x^{2} + y^{2} = 25
<=> / x^{2} + y^{2} = 25
\ 6x + 8y + r^{2} - 50 = 0
The intersection points are on the line with equation
6x + 8y + r^{2} - 50 = 0
This is the equation of AB
The tangent line in M at C_{1} is orthogonal to OM.
The slope of OM is 4/3, so the slope of the tangent line is -3/4
De equation of the tangent line is
y - 4 = (-3/4)(x - 3)
<=> 4y - 16 + 3x - 9 = 0
<=> 3x + 4 y -25 = 0
The tangent line has slope -3/4
The line AB has slope -3/4
They are parallel
A circle C_{1} has center O(0,0) and a circle C_{2} has center D(a,0) with a>0.
Point P(c,d) is an intersection point of the two circles.
The line y = d intersects the circle C_{1} a second time in point A and
intersects the circle C_{2} a second time in point B. Find the distance |A,B|.
Make a clear figure.
The points A and P are located symmetrically with respect to the y-axis.
Therefore, the abscissa of A = x_{A} = -c.
The points B and P are located symmetrically with respect to the line x = a.
Let x_{B} = the abscissa of B. We have
c + x_{B}
--------- = a
2
x_{B} = 2a - c
The distance |A,B| = (2a - c) - (-c) = 2a
Line b has equation x - y = 0
Line c has equation x - 2y = 0
Line d has equation x + 2y = 10
Find the equation of the circle tangent to
the lines b and c and with the center M on the line d.
Point M is in the first quadrant.
M(10 - 2s, s) is a variable point on the line d.
M(10 - 2s, s) with 10-2s > 0 and s > 0 is the center of the circle
<=> (the distance of M to the line b)^{2} = (the distance of M to the line c)^{2}
<=> |M,b|^{2} = |M,c|^{2}
<=> (10 - 3s)^{2} (10 - 4s)^{2}
------------ = ------------ with 10-2s > 0 and s > 0
2 5
<=> ...
<=> M = M(4.095 ; 2.952)
Then we have |M,b|^{2} = 0.653
The requested circle has an equation
(x - 4.095)^{2} + (y - 2.952)^{2} = 0.653
The circle C_{1} has an equation x^{2} + y^{2} = 16.
The circle C_{2} has an equation (x-2)^{2} + y^{2} = 4.
p is a real number between 0 and 4.
The line x = p intersects the circle C_{1} in the points A and B.
Find the value of p such that the circle C_{2} divides the chord [AB] in three equal
parts.
Make a clear figure.
The line x=p intersects the circle C_{1} in a point (p, sqrt(16-p^{2})).
The line x=p intersects the circle C_{2} in a point (p, sqrt( 4 - (p-2)^{2})
Now, we write the condition.
sqrt(16-p^{2}) = 3. sqrt( 4 - (p-2)^{2})
<=> 16-p^{2} = 9.( 4 - (p-2)^{2})
<=> ...
<=> 2 p^{2} - 9p + 4 = 0
The solution for p between 0 and 4 is p=1/2.
Is it possible to draw an isosceles triangle in a circle such that
the area of the triangle is exactly half the area of the circle.
Since the radius of the circle has no influence on the result, we take a
unit circle. The area of this circle is pi.
The largest triangle in a circle is an equilateral triangle.
If the area of this equilateral triangle is smaller than pi/2 , then
it is impossible to draw an isosceles triangle in a circle such that
the area of the triangle is exactly half the area of the circle.
Say, triangle ABC is an equilateral triangle in a unit circle with center O.
The area of the triangle ABC
= 3. (the area of the triangle OAB)
= 3. (1/2) sin(2pi/3)
= 3.sqrt(3)/4 < pi/2
Conclusion: It is impossible to draw an isosceles triangle in a circle such that
the area of the triangle is exactly half the area of the circle.
A Rectangle ABCD is inscribed in a unit circle. BD is a diameter of the circle.
The area of the rectangle is exactly half the area of the circle.
Find the the sizes of the rectangle.
First method
The area of the rectangle should be pi/2
t is the angle BDC
|BC| = 2 sin(t) and |DC| = 2 cos(t)
|BC|.|DC| = 4 sin(t) cos(t) = 2 sin(2t)
2 sin(2t) should be pi/2
<=> sin(2t) = pi/4
<=> 2t = 0.9033
<=> t = 0.45167
|BC| = 0.8729 and |DC| = 1.7994
Second method:
|BC|^{2} + |DC|^{2} = 4 en |BC|^{2}.|DC|^{2} = pi^{2}/4
So, we know the sum and the product of the numbers |BC|^{2} and |DC|^{2}
These numbers are the solutions of the quadratic equation
x^{2} - 4 x + pi^{2}/4 = 0
The solutions are [3.23798, 0.762018]
|BC| = 0.8729 and |DC| = 1.7994
Five tangent circles with equal radius are in a square with side = 10 (see figure).
Find the radius of the circles.
Let r = the radius of the circles.
The diagonal of the small squares is r.sqrt(2)
The diagonal of the big square is 10.sqrt(2)
Now, 10.sqrt(2) = 2.r.sqrt(2) + 4.r
So, r = 5.sqrt(2)/(2 + sqrt(2))
Three tangent circles have radius = 5.
There is a elastic band stretched around the circles (see figure).
Find the length of the elastic band.
The blue segments have length 10.
Together, the three red arcs form exactly one circle with length 20.pi.
The total length of the elastic band is 10 + 20.pi.
Note the usefulness of using a geometric analysis
The circle C_{1} has equation x^{2} + y^{2} = 9
The circle C_{2} has equation (x-6)^{2} + y^{2} = 1
Find the equations of the four common tangent lines.
Take a variable point P(3 cos(t), 3 sin(t)) on C_{1}.
The slope of OP is sin(t)/cos(t).
The slope of the tangent line d to C_{1} in P is then -cos(t)/sin(t).
The equation of line d is
(y - 3 sin(t)) = ( -cos(t)/sin(t) )(x - 3 cos(t))
<=> y sin(t) - 3 sin^{2}(t) = -x cos(t) + 3 cos^{2}(t)
<=> x cos(t) + y sin(t) - 3 = 0
This line d is tangent to C_{2}
<=> The distance from point (6,0) to the line d is 1
<=> | 6 cos(t) - 3 | = 1
First case : 6 cos(t) - 3 = 1
6 cos(t) - 3 = 1
<=> 6 cos(t) = 4
<=> cos(t) = 2/3 and then is sin(t) = ± sqrt(5)/3 (*)
Second case : 6 cos(t) - 3 = -1
6 cos(t) - 3 = -1
<=> 6 cos(t) = 2
<=> cos(t) = 1/3 and then is sin(t) = ± sqrt(8)/3 (**)
From (*) and (**) it follows that the four tangent lines d are
x . (2/3) ± y . sqrt(5)/3 - 3 = 0
and x . (1/3) ± y . sqrt(8)/3 - 3 = 0
After simplification we get
2 x ± sqrt(5) y - 9 = 0
and x ± sqrt(8) y - 9 = 0
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