
Conic
Sections 


Circle
and Line

Equation of a tangent at a point
of a translated circle

Equation of a tangent at a point
of a circle, examples






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. 









College
algebra contents E 



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