\[(4,2)\]

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 \[(3,3)\]

.