circles presentation

1. CIRCLES
CIRCLES

2. CIRCLE-
The set of all points on a
plane that are equidistant
from a fixed point on the
plane. The fixed point is
called the center, and the
fixed line is called the
radius.

3. 1) Find the equation of the
circle of radius 4 which is
centered at ( 3, -2) then
sketch its graph.
Solution:
(x – h)2 + (y – k)2 = r2
(x – 3)2 + (y –(-2))2 = 42
x2 – 6x + 9 + y2 + 4y + 4 = 16
X2 + y2 – 6x + 4y + 13 – 16 = 0
X2 + y2 – 6x + 4y - 3 = 0

4. 2) Change the equation 2x2 + 2y2 – 8x + 5y – 80 = 0 into
standard form.
Solution:
2( x2 – 4x + 4) + 2(y2 +
5
2
𝑦 +
25
16
) = 80 + 8 +
25
8
2 𝑥 − 2 2
+ 2 𝑦 +
5
4
2
=
640+64+25
8
𝑥 − 2 2
+ 𝑦 +
5
4
2
=
27
4
2

5. 3) Express x2 + y2 + Dx + Ey + F = 0 into center-radius
form.
Solution:
(x2+ Dx +
𝐷2
4
) + (𝑦2 + 𝐸𝑦 +
𝐸2
4
) = -F +
𝐷2
4
+
𝐸2
4
𝑥 +
𝐷
2
2
+ 𝑦 +
𝐸
2
2
=
𝐷2
4
+
𝐸2
4
− F
ℎ = −
𝐷
2
; 𝑘 = −
𝐸
2
, 𝑟2 =
𝐷2
4
+
𝐸2
4
− F

6. A. Write the equation of a circle that satisfies the given
condition. Draw its graph
1) Center:( 3, -4), r = 6 6) Center:( 0, 8), r = 5
2) Center:( 5, -12), r = 13 7) Center:( 0, 3), r = 7
3) Center:(5, -3), r = 6 8) Center:( -1, -6), r = 8
4) Center:( 1, -1), r = 3 9) Center:( (2, 4), r = 6
5) Center:( 10, 5), r = 4 10) Center:( 4, -1), r = 6

9. B). Reduce each equation into center-radius form
1) x2 + y2 – 2x + 4y + 5 = 0 6) x2 + y2 – 6x + 2y + 10 = 0
2) x2 + y2 + 4x – 8y – 5 = 0 7) x2 + y2 + 10y = 0
3) x2 + y2 - x = 0 8) x2 + y2 + 8x + 15 = 0
4) x2 + y2 – 4x – 4y + 9 = 0 9) x2 + y2 + 7x + 5y + 16 = 0
5) x2 + y2- 5x – 3y – 4 = 0 10) x2 + y2 – x + y -2 = 0

12. C. Determine whether the given equation
represents a circle, a point, or a null set.
1) x2 + y2 - 2x + 4y + 5 = 0 6) x2 + y2 – 6x + 2y + 10 = 0
2) x2 + y2 + 4x – 8y – 5 = 0 7) x2 + y2 + 10y = 0
3) x2 + y2 –x = 0 8) x2 + y2 + 8x + 15 = 0
4) x2 + y2 – 4x – 4y + 9 = 0 9) x2 + y2 + 7x + 5y + 16 = 0
5) x2 + y2 - 5x – 3y +
34
4
= 0 0)x2 + y2 + ½ x + ½ y +
13
144
= 0

16. Circles
Determined by
Geometric conditions

17. 1) Find the equation of a circle that passes through the points P(1, -2),
Q(5, 4),R(10, 5).
Using x2 + y2 + Dx + Ey + F = 0
(1, 2): 1 + 4 + D - 2E + F = 0
(5, 4): 25 + 16 + 5D + 4E + F = 0
(10, 5): 100 + 25 + 10D + 5E + F = 0
Solving simultaneously
D= -18, E = 6, F = 25
Answer: x2 + y2 – 18x + 6y + 25 = 0

23. 2) A circle is tangent to the line 2x – y + 1 = 0 at the point (2, 5) and the
center is on the line x + y = 9. Find the equation of the circle.
Since radius is perpendicular to
2x – y + 1 = 0 at their point of
tangency, then the equation of the
radius is:
2x – y + 1 = 0 m = 2
mr = -1/2 (2, 5)
y- 5 = -
1
2
𝑥 − 2 , 𝑡ℎ𝑒𝑛
x + 2y = 12
x + y = 9
Solving simultaneously
X = 6, y = 3
r = 6 − 2 2 + 3 − 5 2
R = 20, therefore
𝑥 − 6 2
+ 𝑦 − 3 2
= 20 or
Simplifying further
X2 + y2 - 12x – 6y + 25 = 0

28. This Photo by Unknown Author is licensed under CC BY-NC-ND

30. 

31. 3) A triangle has its sides on the lines x + 2y – 5 = 0, 2x – y -10= 0,
and 2x + y + 2 = 0.
d2
d1
d3
d2

32. −𝑑3=
𝑥+2𝑦−5
1+ 22
, 𝑑2 =
2𝑥−𝑦−10
− 22+ −1 2
, 𝑑1 =
2𝑥+𝑦+2
22+1
𝑑1 = 𝑑2 𝑑2 = 𝑑3
2𝑥+𝑦+2
5
=
2𝑥−𝑦−10
− 5
2𝑥−𝑦−10
− 5
= -
𝑥+2𝑦−5
5
2x + y + 2=-2x+y+10 2x – y – 10= x + 2y – 5
4x = 8 x – 3y = 5
x = 2 y = -1

34. Solving for r:
r = 𝑑1 =
2𝑥+𝑦+2
22+1
(2, -1)
r =
2 2 −1+2
5
= 5
Therefore:
(x -2)2 + (y+1)2 = 5
2
(x -2)2 + (y+1)2 = 5

35. Find the equation of the circle described below. Write
your answer in general form.
1) The circle is tangent to the line x – y = 2 at the point
(4, 2) and the center is on the x - axis.
2) The circle is tangent to the line 4x + 3y = 4 at the
point (4, -4) and the center is on the line x – y = 7.
3) The circle is tangent to the line x + 2y = 3 at the point
(-1, 2) and the center is on the y – axis.
4) The circle is tangent to the line 5x + y = 3 at the point
(2, -7) and the center is on the line x – 2y = 19
5) The circle is tangent to the line 3x – 4y = 34 at the
point (10, -1) and also tangent to the line 4x + 3y = 12
at the point (3, 0)

36. 6) The circle is tangent to both coordinate axes and
contains the point (6, 3).
7) The circle passes through the points (0, 3), (2, 4), and
(1, 0)
8) The circle passes through the points (0, 0), (0, 5), and
(3, 3).
9) The circle is circumscribed about the triangle whose
vertices are (3, -2), (2, 5), and (-1, 6).
10) The circle is circumscribed about the triangle whose
vertices are (-1, -3), (-2, 4), and (2, 1).
11)The sides of a triangle are along the lines x – 2y = 0,
5x – 2y = 8, and 3x + 2y = 24. Find the equation of the
circle inscribed in the triangle.

39. 12) The sides of a triangle lie on the lines 3x + 4y + 8 = 0,
3x – 4y – 32 = 0, and x = 8. Find the equation of the
circle inscribed in the triangle.
13)The sides of a triangle are on the lines 3x – y – 5 = 0,
x + 3y – 1 = 0, and x – 3y + 7 = 0. Find the equation of
the circle inscribed in the triangle.
14)The sides of a triangle are on the lines 6x + 7y = -11,
2x – 9y = -11, and 9x + 2y = 11. Find the equation of
the circle inscribed in the triangle.