MHB How Do You Find the Equation of a Polar Line for a Circle with Given Points?

[mathmari]

Hey!

Let $K$ be a circle with center $C=(x_0,y_0)$ and radius $r$. For each point $P=(p_1, p_2)$ outside the circle let $g_P$ be the line that passes through the intersection points of the tangent from $P$ at the circle and the circle.

I want to find the equation of the line $g_p$ (polar).

I have done the following:

Let $A=(a_1, a_2)$ and $B=(b_1, b_2)$ be the two intersection points.

Since $g_p$ passes through these two points the line has the equation of the form:
$$g_P : y=\frac{b_2-a_2}{b_1-a_1}(x-a_1)+a_2$$

The points $A,B$ are on the circle, so we have the following two relations:
$$(a_1-x_0)^2+(a_2-y_0)^2=r^2 \\ (b_1-x_0)^2+(b_2-y_0)^2=r^2$$

We also have that the triangle MBP and AMP are both right triangles so we could use the Pythagorian Theorem, or not? (Wondering)

Or do we have to find the equation of the tangent? (Wondering)

 

[skeeter]

Let $K$ be a circle with center $C=(x_0,y_0)$ and radius $r$. For each point $P=(p_1, p_2)$ outside the circle let $g_P$ be the line that passes through the intersection points of the tangent from $P$ at the circle and the circle.

I'm not visualizing this part ... not sure what you mean by "at the circle and the circle"

 

[mathmari]

I'm not visualizing this part ... not sure what you mean by "at the circle and the circle"

When we have a point outside the circle and take the tangent to the circle from that point, there are two tangents, so there are two contct points, one (A) between the one tangent and the circle and the other (B) between the other tangent and the circle, right? (Wondering)
I want to find the equation of the line that passes through $A$ and $B$.

 

[mathmari]

I have the following idea:

The equation that we are looking for is \begin{equation*}y-a_2=\frac{b_2-a_2}{b_1-a_1}(x-a_1) \Rightarrow y=\frac{b_2-a_2}{b_1-a_1}(x-a_1)+a_2\end{equation*}
Since from $A$ and $B$ the tangent passes through we have that \begin{equation*}(x-x_0)(a_1-x_0)+(y-y_0)(a_2-y_0)=r^2 \ \text{ und } \ (x-x_0)(b_1-x_0)+(y-y_0)(b_2-y_0)=r^2\end{equation*}

Since $P$ is on that tangent we have that
\begin{equation*}(p_1-x_0)(a_1-x_0)+(p_2-y_0)(a_2-y_0)=r^2 \ \text{ und } \ (p_1-x_0)(b_1-x_0)+(p_2-y_0)(b_2-y_0)=r^2\end{equation*}

Is everything correct so far? Can we continue from here? Or is that way wrong? (Wondering)

 

[skeeter]

This doesn't solve the whole problem, but may be of some help.

since A and B are points on the circle ...

Let $A = (x_A,y_A) = (r_0 \cos{\alpha}, r_0 \sin{\alpha})$ and $B = (x_B,y_B) = (r_0 \cos{\beta}, r_0 \sin{\beta})$, where $r_0$ is the fixed radius of the circle in question and the angles $\alpha$ and $\beta$ are the measure of the rays in standard position that pass thru A and B respectively.

The slope between the two points is $m = \dfrac{\sin{\alpha}-\sin{\beta}}{\cos{\alpha}-\cos{\beta}}$equation of the line through points A and B ...

$y - y_A = m(x-x_A)$

note $x = r\cos{\theta}$ and $y = r\sin{\theta}$

$r\sin{\theta} - r_0 \sin{\alpha} = m(r\cos{\theta} - r_0 \cos{\alpha})$

solving for $r$ as a function of $\theta$ ...

$r = \dfrac{r_0(\sin{\alpha}-m\cos{\alpha})}{\sin{\theta} - m\cos{\theta}}$

I tested the polar function with $r_0=3$, $\alpha=\dfrac{\pi}{3}$, and $\beta=\dfrac{\pi}{6}$

attached polar graphs show that it works ...

 

[Opalg]

Hey!

Let $K$ be a circle with center $C=(x_0,y_0)$ and radius $r$. For each point $P=(p_1, p_2)$ outside the circle let $g_P$ be the line that passes through the intersection points of the tangent from $P$ at the circle and the circle.

I want to find the equation of the line $g_p$ (polar).

I have done the following:

Let $A=(a_1, a_2)$ and $B=(b_1, b_2)$ be the two intersection points.

Since $g_p$ passes through these two points the line has the equation of the form:
$$g_P : y=\frac{b_2-a_2}{b_1-a_1}(x-a_1)+a_2$$

The points $A,B$ are on the circle, so we have the following two relations:
$$(a_1-x_0)^2+(a_2-y_0)^2=r^2 \\ (b_1-x_0)^2+(b_2-y_0)^2=r^2$$

We also have that the triangle MBP and AMP are both right triangles so we could use the Pythagorian Theorem, or not? (Wondering)

Or do we have to find the equation of the tangent? (Wondering)

The line CP has gradient $\dfrac{p_2-y_0}{p_1-x_0}$. The line $AB$ is perpendicular to this, so it has gradient $-\dfrac{p_1-x_0}{p_2-y_0}$, and its equation is therefore $(p_1-x_0)x + (p_2-y_0)y = c$ for some constant $c$.

I don't see a slick way to find $c$. It would be good to avoid having to find the coordinates of $A$ and $B$, or to find the equations of the tangents. My guess is that the best route might be to use the fact that the line $AB$ must pass through the midpoint of $AB$. That is a point on the line $CP$, at a distance $\dfrac{r^2}d$ from $C$, where $d = \sqrt{(p_1-x_0)^2 + (p_2-y_0)^2}$ is the distance from $C$ to $P$.

But even that method seems to involve a rather messy computation.

 

[mathmari]

But even that method seems to involve a rather messy computation.

Ah ok. Which way do you suggest? Maybe using vectors? (Wondering)

- - - Updated - - -

This doesn't solve the whole problem, but may be of some help.

since A and B are points on the circle ...

Let $A = (x_A,y_A) = (r_0 \cos{\alpha}, r_0 \sin{\alpha})$ and $B = (x_B,y_B) = (r_0 \cos{\beta}, r_0 \sin{\beta})$, where $r_0$ is the fixed radius of the circle in question and the angles $\alpha$ and $\beta$ are the measure of the rays in standard position that pass thru A and B respectively.

The slope between the two points is $m = \dfrac{\sin{\alpha}-\sin{\beta}}{\cos{\alpha}-\cos{\beta}}$equation of the line through points A and B ...

$y - y_A = m(x-x_A)$

note $x = r\cos{\theta}$ and $y = r\sin{\theta}$

$r\sin{\theta} - r_0 \sin{\alpha} = m(r\cos{\theta} - r_0 \cos{\alpha})$

solving for $r$ as a function of $\theta$ ...

$r = \dfrac{r_0(\sin{\alpha}-m\cos{\alpha})}{\sin{\theta} - m\cos{\theta}}$

I tested the polar function with $r_0=3$, $\alpha=\dfrac{\pi}{3}$, and $\beta=\dfrac{\pi}{6}$

attached polar graphs show that it works ...

Isn't the picture in the following form:

View attachment 6974

 

[I like Serena]

With vectors:
\begin{tikzpicture}[>=stealth]
\coordinate (C) at (0,0) {};
\coordinate (A) at (3,4) {};
\coordinate (B) at (3,-4) {};
\coordinate (P) at (3+16/3,0);
\coordinate (G) at (3,2);
\draw (C) circle (5);
\draw (A) +(0,1) node[above] {$g$} -- (B) -- +(0,-1);
\draw[->, blue, ultra thick] (C) -- (P);
\draw[->, blue, ultra thick] (C) -- (A);
\draw[<->, blue, ultra thick, dashed] (C)+(-4/10,3/10) -- node[above left] {$\frac{\vec{CP}\cdot \vec{CA}}{CA}$} (3-4/10,4+3/10);
\draw[->, blue, ultra thick] (C) -- (B);
\draw[->, red, ultra thick] (C) -- (G) node

{$G$};
\draw[<->, blue, ultra thick, dashed] (C) +(0,-0.25) -- node[below] {$\frac{\vec{CA}\cdot \vec{CP}}{CP}$} (3,-0.25);
\draw[blue] (A) +(-4/5,3/5) -- (P) -- +(4/5,-3/5);
\draw[blue] (B) +(-4/5,-3/5) -- (P) -- +(4/5,3/5);
\fill (P) circle (0.05) node

{$P$};
\fill (A) circle (0.05) node

{$A$};
\fill (B) circle (0.05) node

{$B$};
\fill (C) circle (0.1) node

{$C$};
\end{tikzpicture}

Any point G on the line g has the same scalar projection:
$$\frac{\vec{CG}\cdot \vec{CP}}{CP} = \frac{\vec{CA}\cdot \vec{CP}}{CP} \tag 1$$
And the length of CA is (by scalar projection):
$$CA = \frac{\vec{CP}\cdot \vec{CA}}{CA} = r \tag 2$$
Combining (1) and (2) gives:
$$\vec{CG}\cdot \vec{CP} = r^2 \quad\Rightarrow\quad (x-x_0)(p_1-x_0) + (y-y_0)(p_2-y_0) = r^2$$
which is the equation of the line $g$.​

 

[mathmari]

I understand! (Happy) At the second question we have to show the following:

Three points A, B and C outside K are collinear iff the lines $g_A$, $g_B$ and $g_C$ have a common point or are pairwise parallel. Let $A=(a_1, a_2)$, $B=(b_1, b_2)$ and $C=(c_1, c_2)$. These points are collinear iff the determinant:

$$\begin{vmatrix}a_1 & a_2 & 1 \\ b_1 & b_2 & 1 \\ c_1 & c_2 & 1\end{vmatrix}$$ is equal to $0$. Then I showed that according to http://mathhelpboards.com/linear-abstract-algebra-14/conditions-so-determinant-zero-21898.html#post98889 that
\begin{equation*}\begin{vmatrix}a_1-x_0 & a_2-y_0 & -x_0(a_1-x_0)-y_0(a_2-y_0) \\ b_1-x_0 & b_2-y_0 & -x_0(b_1-x_0)-y_0(b_2-y_0) \\ c_1-x_0 & c_2-y_0 & -x_0(c_1-x_0)-y_0(c_2-y_0)\end{vmatrix} = 0\end{equation*}
How are the points $A$, $B$ and $C$ placed when the lines $g_A$, $g_B$ and $g_C$ are pairwise parallel? (Wondering)

Should it be as follows?

https://www.physicsforums.com/attachments/6975._xfImport