A circle has centre  
\[(4,2)\]
  and radius  
\[\sqrt{2}\]
.
The equation of the circle is  
\[(x-4)^2+(y-2)^2=(\sqrt{2})^2=2 \rightarrow x^2-8x+y^2-4y+20=2\]
  (1)
A line form the origin forms a tangent with the circle. What is the equation of this tangent and where does it meet the circle?
Let the tangent meet the circle at the point  
\[P(x,y)\]
. The line drawn from the centre of the circle to  
\[P\]
  HAS GRADIENT  
\[\frac{y-2}{x-4}\]
  and the line drawn from the origin has gradient  
\[\frac{y}{x}\]
.
These two lines are at right angles so
\[\frac{y-2}{x-4}=- \frac{x}{y} \rightarrow y^2-2y+x^2-4x=0\]
  (2)
(10-(2) gives  
\[-4x-2y+20=2 \rightarrow y=-2x+9\]

Substitute the equation of this line into (1).  
\[x^2-8x+(-2x+9)^2-4(-2x+9)+20=2 \rightarrow 5x^2-36x+63=0\]
.
Then  
\[x=\frac{36 \pm \sqrt{(-36)^2-4 \times 5 \times 63}}{2 \times 5}=\frac{21}{5}, \: 3\]
.
If  
\[x= \frac{21}{5}\]
  then  
\[y=-2 \times \frac{21}{5}+9=\frac{3}{5}\]
.
If  
\[x= 3\]
  then  
\[y=-2 \times 3+9=3\]
.
There are two lines, with gradients  
\[\frac{3/5}{21/5}=\frac{7}{7}\]
  and equation  
\[y=\frac{1}{7}x\]
  which meets the circle at  
\[(21/5.3/5)\]
, and one with gradient  
\[\frac{3}{3}=1\]
  and equation  
\[y=x\]
  which meets the circle at  
\[(3,3)\]
.