# Intersection of a point and absolute value function contained within a circle

I'm attempting some crazy ideas while programming a game and ran into the following math problem that has been bugging me for a few days:

Given a unit circle and a random point $P$ within the circle, what is the equation that maps an absolute value function such as $y = 1 - |1-x|$ so that the left side passes through the origin, the right side passes through the $P$, and the apex of the absolute value function is on the circle? If it helps, I'm only concerned with the upper-right quadrant.

The end result would be an isosceles triangle with side lengths 1 (the radius) that treats the circle as a kind of reflective surface, but the reflection is like that of a horizontal surface (reversing only y, not x.) I figured the circle "height" function as $y = \sqrt{1-x^2}$ but I'm not sure how to use it to create an absolute value function that also passes through $(0,0)$ and $P$. Any help would be appreciated.

• Are you saying you want to find a transformation of the absolute value function $y=|x|$? – Ataraxia Jun 2 '13 at 10:56
• The left side and right side...of what or whom?? – DonAntonio Jun 2 '13 at 10:57
• A picture would be a big help – bubba Jun 2 '13 at 11:36
• Your question is unclear. Can you sktech the "absolute value" function in the plane that contains the origin, the point $P$ and the apex ? – Yves Daoust Apr 18 '16 at 6:37

A general form of all absolute value functions (translated and scaled) is

$$y=a+k|x+b|$$

There are three conditions. The equation must pass through the origin:

$$a=-k|b|$$

The apex must lie on the unit circle:

$$a^2+b^2=1$$

And the equation must pass through $P=(p_x,p_y)$:

$$p_y=a+k|p_x+b|$$

In order to find an absolute value function that satisfies these requirements, you will need to solve these three simultaneous equations for $a$, $b$, and $k$.

• Thank you for your response. I did not think to use the general form of the absolute value function. That would mean a, b, k are functions of P, correct? So the equation I'm looking for is y = fa(px,py) + fk(px,py) * abs(x + fb(px,py)) where fa, fb, and fk are functions that return the appropriate modifiers a, b, and k for the absolute value function from the given P. But can I find a, b, and k independently from each other using only P? – Jeremy Robson Jun 3 '13 at 9:29

The statements in the answer you already have are all correct. Here is a more explicit way to solve the necessary equations, assuming the coordinates of $P$ are $(p_x,p_y)$ and both coordinates are positive.

The apex of the desired function lies on the intersection of the line $y=\frac{p_y}{p_x} x$ and the circle $x^2 + y^2 = 1$, so it is the point $$A = \left(\frac{p_x}{\sqrt{p_x^2 + p_y^2}}, \frac{p_y}{\sqrt{p_x^2 + p_y^2}}\right).$$ The condition that the function passes through the origin determines that for $x < \frac{p_x}{\sqrt{p_x^2 + p_y^2}}$, the function must follow the formula $$y = \frac{p_y}{p_x} x, \tag1$$ and your "reflection" condition requires that for $x > \frac{p_x}{\sqrt{p_x^2 + p_y^2}}$, the function must follow the formula $$y = \frac{p_y}{\sqrt{p_x^2 + p_y^2}} - \frac{p_y}{p_x}\left(x - \frac{p_x}{\sqrt{p_x^2 + p_y^2}}\right).\tag2$$

We can put this all together in one equation that gives the function value for any real number $x$,

$$y = \frac{p_y}{\sqrt{p_x^2 + p_y^2}} - \frac{p_y}{p_x}\left|x - \frac{p_x}{\sqrt{p_x^2 + p_y^2}}\right|.$$

For $x > \frac{p_x}{\sqrt{p_x^2 + p_y^2}}$ this is clearly the same as equation $(2)$, and for $x < \frac{p_x}{\sqrt{p_x^2 + p_y^2}}$ the constant terms cancel, so the function value defined here is then the same as the value given by equation $(1)$; and of course when $x = \frac{p_x}{\sqrt{p_x^2 + p_y^2}}$, $y = \frac{p_y}{\sqrt{p_x^2 + p_y^2}}$.