# Princess' question at Yahoo! Answers: finding lines through the origin and tangent to given circle

#### MarkFL

Staff member
Here is the question:

How to find equations of tangent to a circle?

Find the equations of tangents to

x^2+y^2-6x-2y+9=0

through the origin. Also find their respective points of contact.

thanks
I have posted a link there to this thread so the OP can see my work.

#### MarkFL

Staff member
Hello Princess,

We are given the equation of the circle:

$$\displaystyle x^2+y^2-6x-2y+9=0$$

The family of lines through the origin is given by:

$$\displaystyle y=mx$$ where $$\displaystyle m\in\mathbb{R}$$

Substituting for $y$, we obtain the following quadratic in $x$:

$$\displaystyle x^2+(mx)^2-6x-2(mx)+9=0$$

Arranging in standard form, we have:

$$\displaystyle \left(m^2+1 \right)x^2-2(3+m)x+9=0$$

Since the line $y=mx$ is tangent to the circle, the discriminant must be zero:

$$\displaystyle \left(-2(3+m) \right)^2-4\left(m^2+1 \right)(9)=0$$

$$\displaystyle 9+6m+m^2-9m^2-9=0$$

$$\displaystyle 8m^2-6m=0$$

$$\displaystyle m(4m-3)=0$$

$$\displaystyle m=0,\,\frac{3}{4}$$

Thus, the tangent lines are:

$$\displaystyle y=0$$

$$\displaystyle y=\frac{3}{4}x$$

Here is a plot of the circle and the two tangent lines:

To find the contact points, we will substitute for $y$ into the circle:

i) $$\displaystyle y=0$$

$$\displaystyle x^2+0^2-6x-2(0)+9=0$$

$$\displaystyle (x-3)^2=0$$

$$\displaystyle x=3$$

This contact point is $$\displaystyle (3,0)$$.

ii) $$\displaystyle y=\frac{3}{4}x$$

$$\displaystyle x^2+\left(\frac{3}{4}x \right)^2-6x-2\left(\frac{3}{4}x \right)+9=0$$

$$\displaystyle \frac{25}{16}x^2-\frac{15}{2}x+9=0$$

$$\displaystyle 25x^2-120x+144=0$$

$$\displaystyle (5x-12)^2=0$$

$$\displaystyle x=\frac{12}{5}\implies y=\frac{9}{5}$$

This contact point is $$\displaystyle \left(\frac{12}{5},\frac{9}{5} \right)$$.