Finding the polar of a point in the unit circle

Question

I wanted to ask about finding the polar of a point in the unit circle.

From the way I was taught, I would find the distance of the point from the origin which is $5$. The inverse of this would be $1/5$. This implies that the inverse point (inverse of $P$) must be at distance $1/5$ from the origin. Using the gradient formula and perpendicular gradient formula, the gradient for the polar is $-4/3$.

From this point onwards, I am unsure of how to continue as all I need is the inverse point in order for me to find the equation of the polar using the point-gradient formula. My teacher uses the distance to the inverse point to find the point however, I am unsure how he does it. I have tried using the distance formula however, at the point(1), (my working out), I have tried some coordinates that have provided the wrong solution. One pair of coordinates I used which satisfied the condition at $(1)$ was $(\sqrt{2}/10 ,\sqrt{2}/10 )$ but, this gave me the wrong answer.

How would one go about finding the inverse point? Thank you and sorry for any formatting errors.

Answer 1

There is very simple way to obtain the polar line of a point $P=(x_0,y_0)$ with respect to the unit circle $x^2+y^2-1=0$ : it is the straight line whose equation is

$$xx_0+yy_0-1=0$$

In your case $P=(4,3)$, it gives :

$$(L) : \ \ 4x+3y-1=0 \ \iff \ y=-\frac43x+\frac13$$

You can check that

• $OP'.OP=\frac15 5 = 1$.

• line (L) intersects the circle in points $T_1,T_2$ which are the tangency points of the tangent lines issued from $P$ to the circle.

Answer 2

There is an analytical way to get this.

Use homogeneous coordinates to describe

• The point $$ \boldsymbol{P} = \pmatrix{4\\3\\1} $$

  A point $(x,y)$ is defined by ${\rm point}(x,y) = \pmatrix{x\\y\\1}$ and the coordinates of $\boldsymbol{P}$ are $(\boldsymbol{P}_1/\boldsymbol{P}_3,\,\boldsymbol{P}_2/\boldsymbol{P}_3)$

• The unit circle $$ {\rm C} = \pmatrix{-1 & & \\ & -1 & \\ & & 1} $$

  Note that the equation of the circle is given by $ \pmatrix{x\\y\\1}^\intercal {\rm C} \pmatrix{x\\y\\1} =0$ which evaluates to $-x^2 - y^2 + 1 =0$

• The polar line of $\boldsymbol{P}$ is evaluated by $$\boldsymbol{L} = {\rm C} \boldsymbol{P}$$

  The equation of a line is $\pmatrix{x \\ y \\1}^\intercal {\rm C}=0$ which evaluates to $(-4)x + (-3)y + 1 =0$ or $y = \frac{1-4 x}{3}$

• A line from the origin to $\boldsymbol{P}$ is defined by $$ \boldsymbol{M} = \pmatrix{0\\0\\1} \times \boldsymbol{L}$$

  With 2D homogenous coordinates the line between two points is calculated with the 3D cross product. In this case $\boldsymbol{M} = \pmatrix{-3 \\ 4 \\ 0}$

• The point $\boldsymbol{Q}$ where the polar line $\boldsymbol{L}$ and the ray $\boldsymbol{M}$ meet is defined by $$ \boldsymbol{Q} = \boldsymbol{L} \times \boldsymbol{M}$$

  The point where two lines meet is also calculated with the cross product and in this case $\boldsymbol{Q} = \pmatrix{-4 \\ -3 \\ -25}$

• The coordinates of the polar point are valued from $\boldsymbol{Q}$

  $$\pmatrix{x \\ y} = \pmatrix{\boldsymbol{Q}_1/\boldsymbol{Q}_3 \\ \boldsymbol{Q}_2 / \boldsymbol{Q}_3} = \pmatrix{ \tfrac{4}{25} \\ \tfrac{3}{25} } $$

• Plot all the points and lines for confirmation