
Conic
Sections 


Circle
and Line

Condition of tangency  Condition for a line to be the tangent to a circle

Condition for a line to be the tangent to the circle with
the center at the
origin
O(0, 0)

Condition for a line to be the tangent to the translated circle

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






Condition of tangency  Condition for a line to be the tangent to a circle

Condition for a line to be the tangent to the circle with center at the
origin
O (0, 0)

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
x^{2}
+ y^{2} = r^{2}
from a line y =
mx + c
or mx
+ y 
c
= 0, 

and after squaring
obtained is 
r^{2}·(m^{2}
+ 1)
= c^{2} 
the condition for a line
y =
mx + c
to be a tangent to 

the circle
x^{2}
+ y^{2}
= r^{2}. 

Condition for a line to be the tangent to the translated circle (x

p)^{2} + (y 
q)^{2} = r^{2} 
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 

after squaring
obtained is 
r^{2}·(m^{2}
+ 1)
= (q m
p 
c)^{2} 


the condition for a line
y =
mx + c
to be a tangent
to a translated circle with the center at S(p,
q). 

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 x^{2}
+ y^{2}
= r^{2}, 
(1) y =
mx + c
plugging (1)
into (2)
gives the quadratic equation, 
(2) x^{2}
+ y^{2} = r^{2
}(m^{2}
+ 1) · x^{2}
+ 2mc · x
+ c^{2}

r^{2}
= 0 and
according to the condition, 

D
= b^{2}

4ac = 0 or
(mc)^{2}
 4(m^{2}
+ 1)(c^{2}
 r^{2})
= 0 gives
r^{2}·(m^{2}
+ 1)
= c^{2}^{
}the tangency condition. 

2)
The condition for a line y =
mx + c
to be tangent to the circle (x

p)^{2} + (y 
q)^{2} = r^{2}, 
(1) y =
mx + c 
(2) (x

p)^{2} + (y 
q)^{2} = r^{2} 

by
plugging (1)
into (2)
obtained is (m^{2}
+ 1) · x^{2}
+ 2(mc  mq
 p)
· x
+ p^{2}
+ c^{2}
+ q^{2} 
r^{2}

2cq = 0 
then,
from the condition D
= b^{2}

4ac = 0
follows
r^{2}·(m^{2}
+ 1)
= (q m
p 
c)^{2}^{
}the
tangency condition. 

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
the 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, 









Conic
sections contents 



Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 