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| Conic
Sections |
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Circle
and Line
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Line circle intersection
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Equation of a tangent at a point
of a circle with the center at the origin
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Equation of a tangent at a point of a translated circle
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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: |
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| If the distance of the center of a circle from a line is such that: |
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d
< r,
then the line intersects the circle in two points, |
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d
= r,
the line touches the circle at only one
point D, |
|
d
> r,
the line does not intersect the circle, and they have no common points. |
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| Equation of a tangent at a point
of a circle with the center at the origin
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| The direction vector of the tangent at the point P1
of a circle and the radius vector of P1
are perpendicular to each other so their scalar product is zero. |
| Points,
O,
P1
and P
in the right figure, determine vectors, |
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| or written in components |
x1x
+ y1y = r2 |
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| This is
equation of a tangent at the point P1(x1,
y1)
of a circle with the center at the origin. |
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| Equation of a tangent at a point
of a translated circle
(x
-
p)2 + (y -
q)2 = r2 |
| The direction vector of the tangent at the point P1(x1, y1), of a circle whose center is at the point
S(p,
q), and the direction vector of the normal, are perpendicular, so their scalar product is zero. |
| Points,
O,
S,
P1
and P
in the right figure, determine vectors, |
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| Since |
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their scalar product is zero, that is |
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| Therefore, |
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is
vector equation of a tangent at the point of a translated
circle, |
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| or,
when this scalar product is written in component form, |
(x1
-
p) · (x
-
p) + (y1
-
q)
· (y
-
q) = r2 |
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| it
represents the equation of the tangent at the point P1
(x1, y1), of a circle whose center is at
S(p,
q). |
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Example:
Find the angle formed by tangents drawn at points of intersection of a line
x -
y + 2 = 0 and
the circle x2
+ y2 = 10. |
| Solution:
Solution of the system of equations gives coordinates of the intersection points, |
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| Plug coordinates of
A
and B
into equation of the tangent: |
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| Example:
At intersections of a line x -
5y + 6 = 0 and the circle
x2
+ y2 -
4x + 2y
-
8 = 0 drown are tangents,
find the area of the triangle formed by the line and the tangents. |
| Solution:
Intersections of the line and the circle are also tangency points. Solutions of the system of
equations are coordinates of the tangency points, |
| (1)
x -
5y + 6 = 0 =>
x =
5y -
6 =>
(2) |
| (2)
x2
+ y2 -
4x + 2y
-
8 = 0 |
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| (5y
- 6)2
+ y2 -
4(5y - 6) + 2y
-
8 = 0 |
| y2 -
3y + 2 = 0,
Þ
y1 =
1 and y2
=
2 |
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x1 =
5 · 1 - 6
= -1,
=>
A(-1,
1), |
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x2 =
5 · 2 - 6
=
4,
=>
B(4,
2). |
| Rewrite
the equation of the circle into standard form, |
| (x
-
p)2 + (y -
q)2 = r2 |
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x2
+ y2 -
4x + 2y
-
8 = 0
=>
(x
-
2)2 + (y +
1)2 = 13, thus
S(2,
-1) and
r = Ö13. |
| Plug
tangency points A and
B
into equation of the tangent, |
| A( -1,
1) and B(4,
2) =>
(x1
-
p) · (x
-
p) + (y1
-
q)
· (y
-
q) = r2 |
| (-1
-
2) · (x
- 2) + (1 +
1)
· (y
+
1) =
13
=> t1
::
-
3x + 2y -
5 = 0, |
| (4
-
2) · (x
- 2) + (2 +
1)
· (y
+
1) =
13
=>
t2
::
2x +
3y -
14 = 0. |
| Solution of the system of equations of tangents determines the third vertex
C
of the triangle, |
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-
3x + 2y -
5 = 0
Tangents are perpendicular
since their slopes satisfy the condition, m1
=
- 1/m2. |
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2x + 3y -
14 = 0,
C(1, 4) |
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| The triangle
ABC is right isosceles, whose area
A =
1/2 · AC 2
=
1/2 · Ö(22
+ 32)2
=
13/2 square units. |
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