
Conic
Sections 


Circle
and Line

Line circle intersection






Circle
and Line

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} = r^{2 }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. 




Example:
At which points the line
x +
5y + 16 = 0
intersects the circle x^{2}
+ y^{2} 
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). 

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. 

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 s_{n
} is perpendicular to the direction
vector s_{t
}.
Therefore, as m_{t}
= s_{y}/s_{x} = 8/15
then 




so, equation of the normal is 

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 


therefore
the tangency point D(7,
5). The radius of the
circle, since 

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. 








Intermediate
algebra contents 



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