Intersection of a Tangent Line From the Origin With a Circle

A circle has centre
$(4,2)$
$\sqrt{2}$
.
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
$P$
$\frac{y-2}{x-4}$
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
$(21/5.3/5)$
$\frac{3}{3}=1$
$y=x$
$(3,3)$ 