P(x,y) is point of C <=> |P,M| = r <=> _______________________ V (x - a)2 + (y - b)2 = r <=> (x - a)2 + (y - b)2 = r2 (1)The last equation is the equation of the circle C.
The equation of the circle with center M(a,b) and radius r is
(x - a)2 + (y - b)2 = r2
x2 + y2 - 2 a x - 2 b y + c = 0 (2)Thus, each circle has an equation of the form (2).
Special Case :
Each circle with center O(0,0) has an equation of the form
x2 + y2 = r2Example :
We calculate the equation of the circle with center A(1,2) and through point B(4,6).
The radius of the circle is |A,B| = sqrt((4-1)2 + (6-2)2 ) = 5.
The equation is (x-1)2 + (y-2)2 = 25
This equation is equivalent with x2 +y2 - 2x - 4y - 20 = 0
Each equation of the form
x2 + y2 + 2 m x + 2 n y + c = 0 (3) with m2 + n2 - c > 0is the equation of a circle.
x2 + y2 + 2 m x + 2 n y + c + m2 + n2 = m2 + n2 <=> (x + m)2 + (y + n)2 = m2 + n2 - c If m2 + n2 - c > 0 then we can write m2 + n2 - c = r2The equation (3) can be written in the form
(x + m)2 + (y + n)2 = r2 So, it is the equation of the circle with center M(-m,-n) ____________ | 2 2 and radius r = \| m + n - c
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 = r2 <=> P(x,y) is on the circle with center (a,b) and radius rConclusion: The circle C with center (a,b) and radius r has parametric equations
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 to such that cos(to)=0.6 and sin(to)= 0.8. From this we have to = 0.927295218
For the point A the t-value is to+pi/3 and then A = A(1.036 , 10.598)
For the point B the t-value is to-pi/3 and then B = B(7.964 , 5.402)
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 x2 + 50 x - 120 = 0 <=> x = 4 or x = 6 Then y = 6 or y = 2The intersection points are (6,2) ; (4,6)
C1 : (x - 2)2 + (y - 5)2 = 25 (6) C2 : (x - 6)2 + (y - 13)2 = 65 (7) The equations are equivalent with / | x2 + y2 - 4 x - 10 y + 4 = 0 | \ x2 + y2 - 12 x - 26 y + 140 = 0 <=> / | x2 + y2 - 4 x - 10 y + 4 = 0 | \ 16 y + 8 x - 136 = 0 <=> / | x2 + y2 - 4 x - 10 y + 4 = 0 | \ x = 17 - 2 y <=> ...The intersection points are (-1,9) and (7,5).
C: (x - a)2 + (y - b)2 = r2 and point P(xo,yo) is on C. b - yo MP has slope ------ a - xo a - xo The tangent line has slope - ------ b - yo The tangent line has equation a - xo y - yo = - ------ (x - xo) b - yoExample :
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).
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
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
t (pi r2) . ------ = (1/2) t r2 2 pi
A circle has radius r.|
The area of the sector with a central angle of t radians is (1/2) t r2
More complex analytic geometry about the circle can be found on a second page about the circle.
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 = x2 and the circle with parametric
The center M(3,4) of the circle C2 is on the circle C1 with
center O(0,0). The circle C2 has a radius r < 10.
The circles have common points A and B.
A circle C1 has center O(0,0) and a circle C2 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 C1 a second time in point A and intersects the circle C2 a second time in point B. Find the distance |A,B|.
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.
The circle C1 has an equation x2 + y2 = 16.|
The circle C2 has an equation (x-2)2 + y2 = 4.
p is a real number between 0 and 4.
The line x = p intersects the circle C1 in the points A and B.
Find the value of p such that the circle C2 divides the chord [AB] in three equal parts.
|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.|
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.
Five tangent circles with equal radius are in a square with side = 10 (see figure).
Find the radius of the circles.
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 circle C1 has equation x2 + y2 = 9|
The circle C2 has equation (x-6)2 + y2 = 1
Find the equations of the four common tangent lines.