How to Calculate the Equation of a Tangent Line to a Circle?

[mathmari]

Hey!

Let $K$ be a circle with center $M=(x_0 \mid y_0)$ and radius $r$ and let $P_1=(x_1\mid y_1)$ be a point of the circle.

I have done the following tofind the equation of the tangent that passes through $P_1$:
The tangent passes through $P_1$ and is perpendicular to $MP$, then let $m_T$ be the slope of the tangent and $m_{MP}$ be the slope of $MP$, then it holds that $m_T\cdot m_{MP}=-1$.

The equation of the tangent is in the form $y=m_Tx+n$.

The slope of $MP$ is $\frac{y_1-y_0}{x_1-x_0}$. The slope of the tangent is therefore \begin{equation*}m_T=-\frac{1}{m_{MP}}=-\frac{1}{\frac{y_1-y_0}{x_1-x_0}}=-\frac{x_1-x_0}{y_1-y_0}\end{equation*}

We get that \begin{equation*}y=-\frac{x_1-x_0}{y_1-y_0}x+n\end{equation*}

Since the tangent passes through $P_1=(x_1\mid y_1)$, it satisfies the equation of the tangent, so we have the following: \begin{equation*}y_1=-\frac{x_1-x_0}{y_1-y_0}x_1+n \Rightarrow n=y_1+\frac{x_1-x_0}{y_1-y_0}x_1\end{equation*}

The equation of the tangent is therefore \begin{align*}&y=-\frac{x_1-x_0}{y_1-y_0}x+y_1+\frac{x_1-x_0}{y_1-y_0}x_1 \\ & \Rightarrow (y_1-y_0)y=-(x_1-x_0)x+y_1(y_1-y_0)+(x_1-x_0)x_1 \\ & \Rightarrow (y_1-y_0)y=-(x_1-x_0)x+y_1(y_1-y_0)-y_0(y_1-y_0)+y_0(y_1-y_0)+(x_1-x_0)x_1-(x_1-x_0)x_0+(x_1-x_0)x_0 \\ & \Rightarrow (y_1-y_0)y=-(x_1-x_0)x+(y_1-y_0)(y_1-y_0)+y_0(y_1-y_0)+(x_1-x_0)(x_1-x_0)+(x_1-x_0)x_0 \\ & \Rightarrow (y_1-y_0)y=-(x_1-x_0)x+(y_1-y_0)^2+y_0(y_1-y_0)+(x_1-x_0)^2+(x_1-x_0)x_0 \\ & \Rightarrow (y_1-y_0)y=-(x_1-x_0)(x-x_0)+(y_1-y_0)^2+y_0(y_1-y_0)+(x_1-x_0)^2 \\ & \Rightarrow (y_1-y_0)y-y_0(y_1-y_0)+(x_1-x_0)(x-x_0)=(y_1-y_0)^2+(x_1-x_0)^2 \\ & \Rightarrow (y_1-y_0)(y-y_0)+(x_1-x_0)(x-x_0)=(y_1-y_0)^2+(x_1-x_0)^2\end{align*}

The equation of the circle is $(x-x_0)^2+(y-y_0)^2=r^2$. Since $P_1$ is a point of the circle we have that $(x_1-x_0)^2+(y_1-y_0)^2=r^2$.

So we get the following equation of the tangent: \begin{equation*}(y_1-y_0)(y-y_0)+(x_1-x_0)(x-x_0)=r^2\end{equation*}
Is everything correct? (Wondering) Let $K$ be a circle with center $M(x_0\mid y_0)$ and radius $r$. For each point $P$ outside the circle, let $g_P$ be the line through the two intersection points of the tangent to $K$ by $O$.

What do we do in this case where $P$ is outside the circle to determine the equation of the tangent? Could you give me a hint?

 

[I like Serena]

Hey mathmari! (Smile)

So we get the following equation of the tangent: \begin{equation*}(y_1-y_0)(y-y_0)+(x_1-x_0)(x-x_0)=r^2\end{equation*}

Is everything correct? (Wondering)

Yep. All correct. (Nod)

Let $K$ be a circle with center $M(x_0\mid y_0)$ and radius $r$. For each point $P$ outside the circle, let $g_P$ be the line through the two intersection points of the tangent to $K$ by $O$.

What do we do in this case where $P$ is outside the circle to determine the equation of the tangent? Could you give me a hint?

Let's consider an alternative for the first part first.

We can express the equation of the tangent as:
$$(\mathbf x - \mathbf x_0)\cdot \mathbf n = r \tag 1$$
where $\mathbf x$ is a vector on the tangent line, $\mathbf x_0 = (x_0 \mid y_0)$, and $\mathbf n$ is the normal vector of unit length.
That is, the distance of the tangent line to the center must be $r$.
Or put otherwise, the projection of the vector from the center to a point on the tangent onto the normal vector of unit length must have length $r$.

In our particular case, we can pick:
$$\mathbf n = \frac{\mathbf x_1 - \mathbf x_0}{\|\mathbf x_1 - \mathbf x_0\|} = \frac{\mathbf x_1 - \mathbf x_0}{r}$$
where $\mathbf x_1 = (x_1 \mid y_1)$

When we substitute this $\mathbf n$ in $(1)$, it follows that:
$$(\mathbf x - \mathbf x_0)\cdot \frac{\mathbf x_1 - \mathbf x_0}{r} = r
\quad\Rightarrow\quad (\mathbf x - \mathbf x_0)\cdot (\mathbf x_1 - \mathbf x_0) = (x-x_0)(x_1-x_0) + (y-y_0)(y_1-y_0) = r^2
$$
(Thinking)Okay. Now let's get back to the second part.
It means that we can again write the equation of the tangent as:
$$(\mathbf x - \mathbf x_0)\cdot \mathbf n = r$$
In particular this holds for $\mathbf P$:
$$(\mathbf P - \mathbf x_0)\cdot \mathbf n = r$$
Can we solve $\mathbf n$ from that? (Wondering)

 

[mathmari]

Let's consider an alternative for the first part first.

We can express the equation of the tangent as:
$$(\mathbf x - \mathbf x_0)\cdot \mathbf n = r \tag 1$$
where $\mathbf x$ is a vector on the tangent line, $\mathbf x_0 = (x_0 \mid y_0)$, and $\mathbf n$ is the normal vector of unit length.
That is, the distance of the tangent line to the center must be $r$.
Or put otherwise, the projection of the vector from the center to a point on the tangent onto the normal vector of unit length must have length $r$.

I haven't really understood why we can write equation of the tangent as $(\mathbf x - \mathbf x_0)\cdot \mathbf n = r$.
The difference $\mathbf x - \mathbf x_0$ is equal to $ \mathbf x_0\mathbf x$, right? Why do we multiply that by the normal vector of unit length to get the radius? (Wondering)

 

[I like Serena]

I haven't really understood why we can write equation of the tangent as $(\mathbf x - \mathbf x_0)\cdot \mathbf n = r$.
The difference $\mathbf x - \mathbf x_0$ is equal to $ \mathbf x_0\mathbf x$, right? Why do we multiply that by the normal vector of unit length to get the radius? (Wondering)

Let's first take a look at what a dot product is.
And in particular how it related to projection.
View attachment 6972
See this article about scalar projection.

More specifically, we can find the projection function $\boldsymbol\pi_{\mathbf n}(\mathbf x)$ that projects a vector $\mathbf x$ onto a vector $\mathbf n$.
\begin{tikzpicture}[>=stealth]
\draw[->, green!70!black, thick] (0,0) -- (4,0) node[below] {$\boldsymbol\pi_{\mathbf n}(\mathbf x)= (\mathbf x\cdot \mathbf n)\mathbf n$};
\draw[->, blue, ultra thick] (0,0) -- node[above left] {$\mathbf x$} (4,3);
\draw[->, red, ultra thick] (0,0) -- node[below] {$\mathbf n$} (1,0);
\draw[green!70!black] (4,0) rectangle +(-0.3,0.3);
\draw[green!70!black, thick, dashed] (4,3) -- (4,0) rectangle +(-0.3,0.3);
\end{tikzpicture}

The projection function is given by:
$$\boldsymbol\pi_{\mathbf n}(\mathbf x) = \frac{\mathbf x\cdot\mathbf n}{\mathbf n \cdot \mathbf n}\ \mathbf n \tag 1$$
In the case that $\mathbf n$ is a vector of unit length, we can simplify this to:
$$\boldsymbol\pi_{\mathbf n}(\mathbf x)= (\mathbf x\cdot \mathbf n)\mathbf n \tag 2$$Are you with me so far? (Wondering)
Next is how we can use this in combination with the Hesse normal form of a line.