Hello Princess,
We are given the equation of the circle:
$$x^2+y^2-6x-2y+9=0$$
The family of lines through the origin is given by:
$$y=mx$$ where $$m\in\mathbb{R}$$
Substituting for $y$, we obtain the following quadratic in $x$:
$$x^2+(mx)^2-6x-2(mx)+9=0$$
Arranging in standard form, we have:
$$\left(m^2+1 \right)x^2-2(3+m)x+9=0$$
Since the line $y=mx$ is tangent to the circle, the discriminant must be zero:
$$\left(-2(3+m) \right)^2-4\left(m^2+1 \right)(9)=0$$
$$9+6m+m^2-9m^2-9=0$$
$$8m^2-6m=0$$
$$m(4m-3)=0$$
$$m=0,\,\frac{3}{4}$$
Thus, the tangent lines are:
$$y=0$$
$$y=\frac{3}{4}x$$
Here is a plot of the circle and the two tangent lines:
View attachment 1704
To find the contact points, we will substitute for $y$ into the circle:
i) $$y=0$$
$$x^2+0^2-6x-2(0)+9=0$$
$$(x-3)^2=0$$
$$x=3$$
This contact point is $$(3,0)$$.
ii) $$y=\frac{3}{4}x$$
$$x^2+\left(\frac{3}{4}x \right)^2-6x-2\left(\frac{3}{4}x \right)+9=0$$
$$\frac{25}{16}x^2-\frac{15}{2}x+9=0$$
$$25x^2-120x+144=0$$
$$(5x-12)^2=0$$
$$x=\frac{12}{5}\implies y=\frac{9}{5}$$
This contact point is $$\left(\frac{12}{5},\frac{9}{5} \right)$$.
