Find all points $~X~$ in $~C: x^2+y^2=1~$ such that the vector $~OX~$ its orthogonal to fixed vector equivalent to free vector $~PX~$ where $~P(0,p)$

Question

Find all points $X$ in $C: x^2+y^2=1$ such that the vector $OX$ its orthogonal to fixed vector equivalent to free vector $PX$ where $P(0,p)$ give the conditions on $p$.

I tried the point $X(0.8,0.6)$: it is a point in $C$ and then we have a vector $OX=[0.8,0.6]$. I made a fixed orthogonal vector first, i.e. $OH=[0.6,-0.8]$, and then the free vector have to be $PX=[0.8,-0.8]$. Now $$ P(0,1.4)\text{ because }0.6-p=-0.8 $$ So I reached the conclusion that $p$ have to be $x+y$ where $x$ and $y$ are the coordinates of point $X$, but it doesn't work for all points and I don't know what I'm doing bad, can you help me?

Answer 1

Let $X$ be the point with coordinates $X(x,y)$, then $OX(x,y)$. Since it lies on $(C)$ we must have $x^2 + y^2 = 1$. Then we must find $(x,y)$ such that $x^2 + y^2 = 1$ and $(x,y) \perp (x-0,y-p)$. Since vectors are orthogonal, then the scalar product is zero, hence $x^2 + y(y-p) = x^2 + y^2 - yp = 1 - yp = 0$. Hence $y = \frac{1}{p}$. Then all points satisfying $x^2 + y^2 = 1$ and $(x,y) \perp (x-0,y-p)$ are $(x,\frac{1}{p})$. But since $x^2+y^2 = 1$, we can write $x$ in terms of $y $ or $p$ as $x = \pm \sqrt{1- y^2} = \pm \sqrt{1 - \frac{1}{p^2}}$. So all points are $$(\pm \sqrt{1- \frac{1}{p^2}}, \frac{1}{p})$$

Answer 2

If I understand you correctly, you are given the circle $$x^2+y^2=1$$ and point on the y-axis, that is the point $P=(0,p)$. Now you want to find a point $X$ on the circle, s.t. that the vector $OX$ is orthogonal to the vector $XP$.

To do this, note that any point on the circle can be described in vector form as
$$X=(x,\pm\sqrt{1-x^2}) \text{ for any } x\in [-1,1]$$
With that in mind, we first look at the upper half of the circle, that is $p>1$ to reduce $X$ to $$X=(x,\sqrt{1-x^2})$$ and thus get the vectors $OX$ and the orthogonal one $XP$ as $$OX=(x,\sqrt{1-x^2}) \text{ and } XP=(\sqrt{1-x^2}, -x) $$

Now we need to fulfill the condition that the line that goes through $X$ and has the direction $XP$ hits the point $(0,p)$. In more formal words $$ (x,\sqrt{1-x^2})+t(\sqrt{1-x^2}, -x)=(0,p) $$ or $$ \begin{align} x+t\sqrt{1-x^2}&=0\\ \sqrt{1-x^2}-tx&=p \end{align} $$ $$ \Leftrightarrow \begin{align} t&=\frac{-x}{\sqrt{1-x^2}}\\ p&=\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}} \end{align} $$

Now we rewrite that to get $x$ from $p$: \begin{equation} x=\sqrt{1-\frac{1}{p^2}} \end{equation} which leads with $x^2+y^2=1$ to $$ y=\frac{1}{p} $$

So to sum up, for given point $(0,p)$, the corresponding points on the circle are $X=(\pm \sqrt{1-\frac{1}{p^2}},\frac{1}{p})$