The document provides information about conic sections, specifically circles. It defines a circle as the set of points equidistant from a fixed point, and provides the standard equation (x - h)2 + (y - k)2 = r2, where (h, k) is the circle's center and r is the radius. Several example problems are worked through, finding the equation of circles given properties like specific points or tangency to lines. The concept of a family of circles is introduced, where the circles share a common property like center location. Radical axes are defined as the line perpendicular to the line joining two circle centers.

1. LECTURE UNIT 003
Conic Sections
General Equation of Conics
Ax2 + By2 + Dx + Ey + F = 0
Circles
Set of points which lie at a fixed distance ( ) from fixed point ( ).
By definition of circle, we have CP = r or
y
P (x, y) (x - h)2 + (y - k)2 = r
r
c (h, k) Standard equation of a circle
x
(x - h)2 + (y - k)2 = r2
Where:
r = radius
(h, k) = center of the circle
(x, y) = any point in the circle
1. Find the equation of the circle with given condition: with center at (-2, -3) and passing through (5, 1).
2. Find the equation of a circle of radius of 5 and center (1, 5).
3. Find the equation of the circle with ends of diameter at (-3, -5) and (5, 1).
4. Find the equation of the circle with center at the y-intercept of the line 4x + 3y - 12 = 0 and passing
through the intersection of 3x + y = 6 and 3x - 2y = 10.
5. Show that the equation x2 - 2x + y2 + 6y = -6 represents a circle. Find its center and radius.
6. Find the center and radius whose equation is 4x2 + 4y2 - 8x + 4y + 1 = 0.
Circle Tangent to the Line Ax + By + C = 0
y
c (h, k) Ax + By + C = 0
r |Ah + Bk + C|
x r=
A2 + B2
7. Find the equation of the circle with center at (-1, 5) and tangent to 3x + y - 6 = 0.
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2. The General Equation of a Circle
The equation x2 + y2 + Dx + Ey + F = 0 is the general equation of the circle, where D, E and F are the arbitrary
constants.
To sketch the graph, reduce the equation to its standard form.
2 2
D E D2 + E2
(x + Dx + 2 ) + (y + Ey + 2 ) = -F + 4 4
2 2 2 2
( x+ D ) ( + y+ E ) = D + E - 4F
2 2 4
D2 + E2 - 4F
r2 =
4
Note:
r2 < 0 no graph (imaginary circle)
r2 = 0 single point
r2 > 0 graph is a circle
Sketch the graph:
8. x2 + y2 - 6x + 2y + 1 = 0
9. x2 + y2 - 5x + 3y + 20 = 0
10. 4x2 + 4y2 - 8x + y - 4 = 0
11. x2 + y2 + 10x + 25 = 0
12. x2 + y2 - 8x + 6y - 11 = 0
13. 2x2 + 2y2 - 7x + y + 100 = 0
14. 3x2 + 3y2 - 9x + 3y - 24 = 0
15. x2 + y2 - 2x + 4y + 5 = 0
Circle Determined by Three Conditions
x2 + y2 + Dx + Ey + F = 0 (General equation of a circle)
Where:
D, E and F are the arbitrary constants
Use this equation when there are 3 points in the circle
(x - h)2 + (y - k)2 = r2 (Standard equation of
a circle)
Where:
h, k and r are the arbitrary constants
Use this equation when the given radius of the circle is tangent to a line.
16. Find the equation of the circle passing through (1, -2), (3, 1) and (2, -1).
17. Passing through the origin and the points (5, -2) and (3, 3). Find the equation of the circle.
18. With center on the y=axis and passing through (2, -3) and (-1, 4).
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3. Family of Circles
19. Find the equation of the family of circles with center at (3, -2).
20. Find the equation of the family of circles with center on the x-axis.
To find the the equation of the family of circles passing through the intersection of 2 circles whose equation are:
x2 + y2 + D1x + E1y + F1 = 0 (C1)
x2 + y2 + D2x + E2y + F2 = 0 (C2)
C1 and C2 are non concentric circles
Equation of the family is
x2 + y2 + D1x + E1y + F1 + k(x2 + y2 + D2x + E2y + F2) = 0
Where:
k = 0 Will result to another original equation.
k = -1 Will result to a linear equation.
If k = -1 the resulting equation is the equation of the radical
axis (R.A.) of 2 circles.
Radical axis - a line perpendicular to the line joining the centers of the 2 circles.
y y
R.A.
R.A.
C1 (h1, k1)
C1
x x
C2 (h2, k2)
C2
21. Find the equation of the circle that passes through the intersection of x2 + y2 - 2x + 4y - 4 = 0 and
x2 + y2 + 4x + 6y - 3 = 0. (a) Find the equation of one member (b) find the equation of the radical axis
and (c) find the equation of the member that passes through (-3, 4).
22. Find the equation of the circle that passes through the intersection of x2 + y2 + 3x - y - 5 = 0 and
x2 + y2 - 2x + 4y - 11 = 0 and the point (-1, 2).
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4. CIRCLES
Example 1:
Find the equation of the circle with given condition: with center at (-2, -3) and passing through (5, 1).
Solution:
From; (x - h)2 + (y - k)2 = r2
(5 + 2)2 + (1 + 3)2 = r2
r2 = 65
(5, 1)
r = 65
r = 8.06226
Hence; (-2, -3)
(x + 2)2 + (y + 3)2 = 65
r = 8.06226
(x + 2)2 + (y + 3)2 = 65
Example 5:
Show that the equation x2 - 2x + y2 + 6y = -6 represents a circle. Find its center and radius.
Solution:
x2 - 2x + y2 + 6y = -6
Completing squares:
(x2 - 2x + 1) + (y2+ 6y + 9) = -6 + 1 + 9
(x - 1)2 + (y + 3)2 = 4
C (1, -3)
r=2
Example 7:
Find the equation of the circle with center at (-1, 5) and tangent to 3x + y - 6 = 0.
Solution:
(x + 1)2 + (y - 5)2 = r2
(0, 6)
(-1, 5)
| 3(-1) + 1(5) - 6|
r= 3x + y - 6 = 0
32 + 12
4
r =-
10 (2, 0)
r2 = 8
5
(x + 1)2 + (y - 5)2 = 8
5
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5. Example 8:
Sketch the graph x2 + y2 - 6x + 2y + 1 = 0
Solution:
(x2 - 6x + 9) + (y2 + 2y + 1) = -1 + 1 + 9
(x - 3)2 + (y + 1)2 = 9
C (3, -1) (3, -1)
r=3
r=3
Example 16:
Find the equation of the circle passing through (1, -2), (3, 1) and (2, -1).
Solution:
P1 (1, -2) D - 2E + F = -5 eq. 1
P2 (3, 1) 3D + E + F = -10 eq. 2
P3 (2, -1) 2D - E + F = -5 eq. 3
By elimination and substitution, we find that:
D=5
E = -5
(2.5, 2.5)
F = -20
Therefore: r = 5.7 (3, 1)
x2 + y2 + 5x - 5y - 20 = 0
(2, -1)
(1, -2)
(x + 2.5)2 + (y - 2.5)2 = 65/2
Example 16:
With center on the y=axis and passing through (2, -3) and (-1, 4).
Solution:
Center on the y-axis, C (0, k)
x2 + (y - k)2 = r2 eq. 1
(-1, 4)
(2, -3) 4 + (-3 - k)2 = r2
13 + 6k + k2 = r2 eq. 2
(-1, 4) 1 + (4 - k)2 = r2
17 - 8k + k2 = r2 eq. 3 (0, 2/7)
Solving simultaneously;
k = 2/7
r2 = 725/49
(2, -3)
Required equation:
X2 + ( y - 2 ) = 725
2
7 49
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6. Example 22:
Find the equation of the circle that passes through the intersection of x2 + y2 + 3x - y - 5 = 0 and
x2 + y2 - 2x + 4y - 11 = 0 and the point (-1, 2).
Solution:
x2 + y2 + 3x - y - 5 + k (x2 + y2 - 2x + 4y - 11) = 0 equation of the family
at (-1, 2) we find k = 5/4
Substitute the value of k, and determine the equation of the circle:
X2 + y2 + 3x - y - 5 + (5/4)(x2 + y2 - 2x + 4y - 11) = 0
Solving simultaneously, the desired equation of the circle is:
9x2 + 9y2 + 2x + 16y - 75 = 0
To determine the equation of the radical-axis, k = -1
5x - 5y + 6 = 0
x2 + y2 + 3x - y - 5 = 0
(-1, 2) 5x - 5y + 6 = 0
(-1.5, 0.5)
(-1/9, -8/9)
(1, -2)
9x2 + 9y2 + 2x + 16y - 75 = 0 x2 + y2 - 2x + 4y - 11 = 0
Example 19:
Find the equation of the family of circles with center at (3, -2).
Solution:
(x - 3)2 + (y + 2)2 = r2 equation of the family
r=1
(x - 3)2 + (y + 2)2 = 1 equation of first member
r=3
(x - 3)2 + (y + 2)2 = 9 equation of the second member
Example 20:
Find the equation of the family of circles with center on the x-axis.
Solution:
(x - h)2 + y2 = r2 equation of the family
r = 4, h = -2
(x + 2)2 + y2 = 16 equation of one member
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