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Conic
Sections |
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Circle
and Line
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Condition of tangency - Condition for a line to be the tangent to a circle
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Condition for a line to be the tangent to the circle with
the center at the
origin
O(0, 0)
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Condition for a line to be the tangent to the translated circle
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Tangents to a circle from a point outside the circle - use of the
tangency condition
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Condition of tangency - Condition for a line to be the tangent to a circle
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Condition for a line to be the tangent to the circle with center at the
origin
O (0, 0)
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A line touches a circle if the distance of the center of the circle to the line is equal to the radius of the circle,
i.e., if d
= r. |
The distance of the center
S
(0,
0) of a circle
x2
+ y2 = r2
from a line y =
mx + c
or -mx
+ y -
c
= 0, |
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and after squaring
obtained is |
r2·(m2
+ 1)
= c2 |
the condition for a line
y =
mx + c
to be a tangent to |
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the circle
x2
+ y2
= r2. |
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Condition for a line to be the tangent to the translated circle (x
-
p)2 + (y -
q)2 = r2 |
In this case, the distance
d
of the center
S(p,
q) of the circle to a line
-mx
+ y -
c
= 0 (
or
y =
mx + c ),
must be equal to the radius r
of the circle, thus |
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after squaring
obtained is |
r2·(m2
+ 1)
= (q -m
p -
c)2 |
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the condition for a line
y =
mx + c
to be a tangent
to a translated circle with the center at S(p,
q). |
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We can derive the same conditions using
the tangency criteria which implies that the discriminant of the
system of equations of a line and a circle, in that case, must be zero. |
Then will the system have only one solution, i.e., the line and the circle will have only one common point,
the tangency point. |
1)
The condition for a line y =
mx + c
to be tangent to the circle x2
+ y2
= r2, |
(1) y =
mx + c
plugging (1)
into (2)
gives the quadratic equation, |
(2) x2
+ y2 = r2
(m2
+ 1) · x2
+ 2mc · x
+ c2
-
r2
= 0 and
according to the condition, |
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D
= b2
-
4ac = 0 or
(mc)2
- 4(m2
+ 1)(c2
- r2)
= 0 gives
r2·(m2
+ 1)
= c2
the tangency condition. |
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2)
The condition for a line y =
mx + c
to be tangent to the circle (x
-
p)2 + (y -
q)2 = r2, |
(1) y =
mx + c |
(2) (x
-
p)2 + (y -
q)2 = r2 |
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by
plugging (1)
into (2)
obtained is (m2
+ 1) · x2
+ 2(mc - mq
- p)
· x
+ p2
+ c2
+ q2 -
r2
-
2cq = 0 |
then,
from the condition D
= b2
-
4ac = 0
follows
r2·(m2
+ 1)
= (q -m
p -
c)2
the
tangency condition. |
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Tangents to a circle from a point outside the circle - use of the
tangency condition
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Example:
Find the angle between tangents drawn from the point
A(-1,
7) to the circle
x2
+ y2 = 25 . |
Solution:
Equations of tangents we
find from the system formed by equation of the line and the
tangency condition, |
A(-1,
7) => (1)
y =
mx + c => c =
m + 7
=> (2) |
(2) r2·(m2
+ 1)
= c2 |
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25(m2
+ 1)
= (m + 7)2 |
12m2
-
7m
-
12 = 0 => m1
= -
3/4
and
m2
= 4/3 |
c1
= -
3/4 + 7 = 25/4 and
c2
= 4/3 + 7 = 25/3 |
therefore,
the equations of tangents, |
t1
::
y =
-
(3/4)x + 25/4 and
t2
::
y =
(4/3)x + 25/3. |
Slopes of tangents satisfy perpendicularity condition, that is
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m1
= -
1/m2
=>
j =
90°. |
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Example:
Find the area of the triangle made by points of contact of tangents, drawn from the point |
A(15,
12) to the circle
(x
-
5)2 + (y -
2)2 =
20 , and the center
S of the circle. |
Solution:
From the system of equations formed by equation of the line through point
A
and the condition of
tangency for the circle with the center at S(p,
q), we calculate slopes and intersections of tangents, thus |
A(15,
12)
=> (1) y =
mx + c, c =
12 -
15m
=>
(2) |
(2) r2·(m2
+ 1) = (q -m
p -
c)2 |
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S(5,
2)
and
r2
= 20
=>
(2) |
20·(m2
+ 1) = (2 -
5m
-
12 +
15m)2 |
2m2
-
5m
+ 2 = 0
=>
m1
= 1/2
and
m2
= 2, |
as
c =
12 -
15m
then c1
= 9/2 and
c2
= -
18 |
therefore,
the equations of tangents, |
t1
::
y =
(1/2)x + 9/2
and
t2
::
y =
2x -
18. |
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Coordinates of the tangency points we calculate
by solving system of equations formed by equations of tangents and equation of the
circle, thus |
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and
the tangency point D2, |
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The
area of the triangle SD1D2, |
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Example:
Given is a line
-3x +
y + 1 = 0 and
the circle
x2
+ y2 -
6x
- 4y
+ 3 = 0,
find equations of tangents to the circle which are perpendicular to the line. |
Solution:
Slopes of tangents are determined by condition of
perpendicularity, therefore |
y =
3x -
1,
m =
3
so that
mt =
-1/3 |
x2
+ y2 -6x
-4 y
+ 3 = 0
=>
(x
-
3)2 + (y -
2)2 = 10 |
thus,
S(3,
2)
and
r2
= 10. |
To
find intersections c
we use the tangency condition, |
r2·(m2
+ 1)
= (q -m
p -
c)2 |
10[(-1/3)2
+ 1]
= [2 -(-1/3)·3
-
c]2 or
(3
-
c)2
= 100/9 |
(3
-
c)
= ± 10/3
so c1
= -1/3
and c2
= 19/3. |
The equations of
tangents are, |
t1
::
y =
-(1/3)x
-1/3 and t2
::
y =
-(1/3)x
+19/3. |
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Example: Find equations of the common tangents to
circles x2
+ y2 =
13 and (x
+ 2)2 + (y +
10)2 =
117. |
Solution:
Slopes and intersections of common tangents to the circles must satisfy tangency condition of both
circles. Therefore, values for slopes m
and intersections c
we calculate from the system of equations, |
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The
equation -10
+ 2m -
c = +3c
does not satisfy given conditions. |
The
equation -10
+ 2m -
c = -3c
or c = 5 -
m plugged into (1) |
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Therefore,
the equations of the common tangents are, |
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