
Conic
Sections 


Circle
and Line

Line circle intersection

Equation of a tangent at a point
of a circle with the center at the origin

Equation of a tangent at a point of a translated circle







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. 

Equation of a tangent at a point
of a circle with the center at the origin

The direction vector of the tangent at the point P_{1}
of a circle and the radius vector of P_{1}
are perpendicular to each other so their scalar product is zero. 
Points,
O,
P_{1}
and P
in the right figure, determine vectors, 


the
scalar product written in
the components gives,

x_{1}x
+ y_{1}y = r^{2}

This is
equation of a tangent at the point P_{1}(x_{1},
y_{1})
of a circle with the center at the origin. 




Equation of a tangent at a point
of a translated circle
(x

p)^{2} + (y 
q)^{2} = r^{2 } 
The direction vector of the tangent at the point P_{1}(x_{1}, y_{1}), 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,
P_{1}
and P
in the right figure, determine vectors, 

Since 

their scalar product is zero, that is 






Therefore, 

is
vector equation of a tangent at the point of a translated
circle, 

or,
when this scalar product is written in the component form, 
(x_{1}

p) · (x

p) + (y_{1}

q)
· (y

q) = r^{2 } 

it
represents the equation of the tangent at the point P_{1}
(x_{1}, y_{1}), of a circle whose center is at
S(p,
q). 

Example:
Find the angle formed by tangents drawn at points of intersection of a line
x 
y + 2 = 0 and
the circle x^{2}
+ y^{2} = 10. 
Solution:
Solution of the system of equations gives coordinates of the intersection points, 

Plug coordinates of
A
and B
into equation of the tangent: 






Example:
At intersections of a line x 
5y + 6 = 0 and the circle
x^{2}
+ y^{2} 
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)
x^{2}
+ y^{2} 
4x + 2y

8 = 0 

(5y
 6)^{2}
+ y^{2} 
4(5y  6) + 2y

8 = 0 
y^{2} 
3y + 2 = 0,
=> y_{1} =
1 and y_{2}
=
2 
x_{1} = 5 · 1  6
= 1,
=> A(1,
1), 
x_{2} =
5 · 2  6
=
4, => B(4,
2). 
Rewrite
the equation of the circle into standard form, 
(x

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



x^{2}
+ y^{2} 
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) => (x_{1}

p) · (x

p) + (y_{1}

q)
· (y

q) = r^{2 } 
(1

2) · (x
 2) + (1 +
1)
· (y
+
1) =
13
=> t_{1}
:: 
3x + 2y 
5 = 0, 
(4

2) · (x
 2) + (2 +
1)
· (y
+
1) =
13
=> t_{2}
:: 2x +
3y 
14 = 0. 
Solution of the system of equations of tangents determines the third vertex
C
of the triangle, 

3x + 2y 
5 = 0
Tangents are perpendicular
since their slopes satisfy the condition, m_{1}
=
 1/m_{2}. 
2x + 3y 
14 = 0,
C(1, 4) 

The triangle
ABC is right isosceles, whose area
A =
1/2 · AC ^{2}
=
1/2 · Ö(2^{2}
+ 3^{2})^{2}
=
13/2 square units. 








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