### circle

```x2
+
circle
y2 =
100
tangent line
4x – 3y = 50
point of tangency
(8,6)
is the negative reciprocal
to the slope of the
tangent line.
3
4
mt =
mr =
3
4
The circle above is defined by the equation: x2 + y2 = 100.
A tangent line (4x – 3y = 50) intersects the circle at a
point of tangency: (8,6). The tangent line is
perpendicular to the radius of the circle.
When a tangent and a radius intersect at the point of
tangency, they are always perpendicular to each other.
It then follows that their slopes are always negative
reciprocals of each other.
yP  yC
mr 
xP  xC
 1 2
mr 
2  ( 1)
C(-1,2)
3
mr 
 1
3
yP  yt
P(2,-1)
mt 
xP  xt
T(0,-3)
 1  ( 3)
mt 
20
1
mr =
2
mt   1
mt
2
Find the slope of the line tangent to the circle x2 + y2 = 5
and passing through the point R(-2,1).
yR  yC
mr 
xR  xC
1 0
mr 
20
1
mr  
2
1
mt  
mr
 mt  2
R(-2,1)
Find the equation of the tangent to the circle x2 + y2 +
10x – 24y = 0 and passing through the point T(0,0).
Step 1: Find the centre and the radius.
x2 + y2 + 10x – 24y = 0
(x2 + 10x + 25) + (y2 – 24y + 144) = 0 + 25 + 144
(x + 5)2 + (y – 12)2 = 169
Step 3: Find the slope of the tangent.
1
5
Centre: (-5,12)
r = 13
mt  
mt 
mr
12
Step 2: Find the slope of the radius.
yT  yC
Step 4: Find the
y  yT
mr 
equation of the mt 
xT  xC
x  xT
tangent.
0  12
y0
5
mr 

0  ( 5)
12 x  0
 12
12(y – 0) = 5(x – 0)
mr 
5
12y = 5x
5x – 12y = 0
Find the equation of the tangent to the circle x2 + y2 – 6y
- 16 = 0 and passing through the point T(3,7).
Step 1: Find the centre and the radius.
x2 + y2 – 6y - 16 = 0
x2 + (y2 – 6y + 9) = 16 + 9
x2 + (y – 3)2 = 25
Centre: (0,3) r = 5
yT  yC
mr 
xT  xC
73
mr 
30
4
mr 
3
1
mt  
mr
3
mt  
4
y  yT
mt 
x  xT
3 y7
 
4 x3
4(y – 7) = -3(x – 3)
4y - 28 = -3x + 9
3x + 4y = 9 + 28
3x + 4y = 37
```