A circle has centre  {jatex options:inline}(4,2){/jatex}  and radius  {jatex options:inline}\sqrt{2}{/jatex}.
The equation of the circle is  {jatex options:inline}(x-4)^2+(y-2)^2=(\sqrt{2})^2=2 \rightarrow x^2-8x+y^2-4y+20=2{/jatex}  (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  {jatex options:inline}P(x,y){/jatex}. The line drawn from the centre of the circle to  {jatex options:inline}P{/jatex}  HAS GRADIENT  {jatex options:inline}\frac{y-2}{x-4}{/jatex}  and the line drawn from the origin has gradient  {jatex options:inline}\frac{y}{x}{/jatex}.
These two lines are at right angles so
{jatex options:inline}\frac{y-2}{x-4}=- \frac{x}{y} \rightarrow y^2-2y+x^2-4x=0{/jatex}  (2)
(10-(2) gives  {jatex options:inline}-4x-2y+20=2 \rightarrow y=-2x+9{/jatex}
Substitute the equation of this line into (1).  {jatex options:inline}x^2-8x+(-2x+9)^2-4(-2x+9)+20=2 \rightarrow 5x^2-36x+63=0{/jatex}.
Then  {jatex options:inline}x=\frac{36 \pm \sqrt{(-36)^2-4 \times 5 \times 63}}{2 \times 5}=\frac{21}{5}, \: 3{/jatex}.
If  {jatex options:inline}x= \frac{21}{5}{/jatex}  then  {jatex options:inline}y=-2 \times \frac{21}{5}+9=\frac{3}{5}{/jatex}.
If  {jatex options:inline}x= 3{/jatex}  then  {jatex options:inline}y=-2 \times 3+9=3{/jatex}.
There are two lines, with gradients  {jatex options:inline}\frac{3/5}{21/5}=\frac{7}{7}{/jatex}  and equation  {jatex options:inline}y=\frac{1}{7}x{/jatex}  which meets the circle at  {jatex options:inline}(21/5.3/5){/jatex}, and one with gradient  {jatex options:inline}\frac{3}{3}=1{/jatex}  and equation  {jatex options:inline}y=x{/jatex}  which meets the circle at  {jatex options:inline}(3,3){/jatex}.