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| Conic
Sections |
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Circle
and Line
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Line circle intersection
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| Circle
and Line
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| Line circle intersection |
| A
line and a circle in a plane can have one of the three positions in relation
to each other, depending on the distance d
of the center S(p,
q) of
the circle (x
-
p)2 + (y -
q)2 = r2 from
the line Ax
+ By + C = 0, where the formula for the distance: |
 |
| If the distance of the center of a circle from a line is such that: |
|
d
< r,
then the line intersects the circle in two points, |
|
d
= r,
the line touches the circle at only one point, |
|
d
> r,
the line does not intersect the circle, and they have no common points. |
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| Example:
At which points the line
x +
5y + 16 = 0
intersects the circle x2
+ y2 -
4x + 2y -
8 = 0. |
| Solution:
To find coordinates of points at which the line intersects the circle solve the system of equations: |
 |
 |
| So, the line intersects the circle at points,
A(4, -4)
and B(-1,
-3). |
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| Example:
Find equation of a circle with the center at
S(1,
20) which touches the
line 8x +
15y -
19 = 0. |
| Solution:
If a line touches a circle then the distance between the tangency point and the center of the circle |
| d
= DS =
r i.e., |
 |
| thus,
equation of the circle (x
-
1)2 + (y -
20)2 = 289. |
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| We can use another method to solve this problem. Since, the
normal n through the center is perpendicular to the tangent
t
then the direction vector sn
is perpendicular to the direction
vector st
.
Therefore, as mt
= sy/sx = -8/15
then |
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| so, equation of the normal is |
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As the tangency point
D
is the common point of the tangent and the normal then, putting coordinates of the
radius vector of the normal into equation of the tangent determines a value of the parameter
l
to satisfy that
condition, as |
 |
| then,
these variable coordinates of the radius vector put into equation of the tangent |
 |
| follows
8 ·
(1 + 8l) +
15 · (20 + 15l)
-
19 = 0 =>
289l
= 289 => l
= -
1 |
| so, the radius vector of the tangency point |
 |
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| therefore
the tangency point D(-7,
5). The radius of the
circle, since |
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| This result we can check
by plugging the coordinates of the tangency point into equation of the circle that is |
| D(-7,
5) =>
(x
-
1)2 + (y -
20)2 = 289,
(-7
-
1)2 + (5 -
20)2 = 289
=>
(-8)2 + (-
15)2 = 289 |
| therefore, the tangency point is the point of the circle. |
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