note: Everything here is in 2 dimensions
Let $|\vec{r}|$ be constant, so the graph of all possible $\vec{r}$ defines a circle. Let $|\vec{P}| = |P \bullet \hat{y}|$. Let $\vec{D}$ be a vector in the 4th quadrant.
If $\vec{P}$ and $\vec{P}+\vec{D}$ are on the circle, find $|\vec{r}|$ in terms of $\vec{P}$ and $\vec{D}$
I can solve this by going into components of the equation of the circle: $$x^2 + y^2 = r^2$$ and substituting in $\vec{P}+\vec{D}$: $$D_x^2 + (P_y+D_y)^2 = r^2$$ and since $ P_y = r$ $$D_x^2 + (r+D_y)^2 = r^2$$ solving for r to get $$ r = \frac{D_x^2+Dy^2}{-2D_y}$$
where $D_y$ is a negative number.
My question is how you would do this without going into components, and still get the equivalent vector answer $$|\vec{r}| = \frac{|\vec{D}|}{-2\vec{D} \bullet \hat{y}}$$
I suspect you mean that $P$ is parallel to the basis vector $\hat y = (0,1)$. I will drop all "arrow"s i.e. just write $P$ instead of $\vec P$. Let $r$ be the radius of the circle in question. Then $P = r\hat y$. The proof you are looking for comes from expanding the dot product $(P+D)\cdot(P+D) = |P+D|^2$:
$$r^2 = |P + D|^2 = |P|^2 + |D|^2 + 2 P\cdot D = r^2 + |D|^2 + 2rD\cdot \hat y$$ which rearranges to $r = \frac{|D|^2}{-2D\cdot \hat y}$.