

Circle
and Line

Tangents to a circle from a point outside the circle  use of the
tangency condition

Angle between a line and a circle






Tangents to a circle from a point outside the circle  use of the
tangency condition

Example:
Find the angle between tangents drawn from the point
A(1,
7) to the circle
x^{2}
+ y^{2} = 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) r^{2}·(m^{2}
+ 1)
= c^{2} 

25(m^{2}
+ 1)
= (m + 7)^{2} 
12m^{2
}
7m

12 = 0 => m_{1}
= 
3/4
and
m_{2}
= 4/3 
c_{1}
= 
3/4 + 7 = 25/4 and
c_{2}
= 4/3 + 7 = 25/3 
therefore,
the equations of tangents, 
t_{1}
::
y =

(3/4)x + 25/4 and
t_{2}
::
y =
(4/3)x + 25/3. 
Slopes of tangents satisfy perpendicularity condition, that is

m_{1}
= 
1/m_{2}
=>
j =
90°. 




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) r^{2}·(m^{2}
+ 1) = (q m
p 
c)^{2} 

S(5,
2)
and
r^{2}
= 20
=>
(2) 
20·(m^{2}
+ 1) = (2 
5m

12 +
15m)^{2} 
2m^{2
}
5m
+ 2 = 0
=>
m_{1}
= 1/2
and
m_{2}
= 2, 
as
c =
12 
15m
then c_{1}
= 9/2 and
c_{2}
= 
18 
therefore,
the equations of tangents, 
t_{1}
::
y =
(1/2)x + 9/2
and
t_{2}
::
y =
2x 
18. 



Coordinates of the tangency points we calculate
by solving system of equations formed by equations of tangents and equation of the
circle, thus 

and
the tangency point D_{2}, 

The
area of the triangle SD_{1}D_{2}, 


Example:
Given is a line
3x +
y + 1 = 0 and a circle
x^{2}
+ y^{2} 
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
m_{t} =
1/3 
x^{2}
+ y^{2} 6x
4 y
+ 3 = 0
=>
(x

3)^{2} + (y 
2)^{2} = 10 
thus,
S(3,
2)
and
r^{2}
= 10. 
To
find intersections c
we use the tangency condition, 
r^{2}·(m^{2}
+ 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 c_{1}
= 1/3
and c_{2}
= 19/3. 
The equations of
tangents are, 
t_{1}
::
y =
(1/3)x
1/3 and t_{2}
::
y =
(1/3)x
+19/3. 




Example: Find equations of the common tangents to
circles x^{2}
+ y^{2} =
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, 

The
equation 10
+ 2m 
c = +3c
does not satisfy given conditions. 
The
equation 10
+ 2m 
c = 3c
or c = 5 
m plugged into (1) 

Therefore,
the equations of the common tangents are, 


Angle between a line and a circle

The angle between a line and a circle is the angle formed by the line and the tangent to the circle at the
intersection point of the circle and the given line. 

Example:
Find the angle between a line
2x +
3y 
1 = 0 and a circle
x^{2}
+ y^{2} +
4x
+ 2y  15
= 0. 
Solution:
Coordinates of
intersections of the line and the circle calculate by solving the system, 

Rewrite
the equation of the circle to standard form, 
x^{2}
+ y^{2} +
4x
+ 2y  15
= 0
=>
(x

p)^{2} + (y 
q)^{2} = r^{2} 
thus,
(x +
2)^{2} + (y + 1)^{2} = 20,
S(2,
1) and
r^{2}
= 20. 



Equation of the tangent at the intersection
S_{1}, 
S_{1}(4,
3)
=>
(x_{1} 
p) ·
(x 
p) + (y_{1} 
q) ·
(y 
q) = r^{2}

(4
+
2)
· (x +
2)
+ (3 + 1)
·
(y
+
1)
= 20
=>
t
::

x
+ 2y

10
= 0
or y =
1/2x + 5 
The angle between the line and the circle is the angle formed by the line and the tangent to the circle at the
intersection point, therefore










Precalculus contents
H 



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