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I have a Point P in unit circle (on or in it) with a radius of r. How can I calculate a Point Q with a fixed radius of x, which has the same angle like P

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Do you mean the point $Q$ lies on the same ray from the origin as $P$, just with a different distance from the origin? – Suugaku May 21 '13 at 16:31

Let $P=(x_1, y_1)$ and $Q=(x_2,y_2)$, for unknown $(x_2,y_2)$. We want the points to be in the same line connecting the origin $(0,0)$ and $P$. (Let's assumme $P$ is not the origin!) This line is given by $$ x_1 y = y_1 x $$ Note that I didn't write $y=\frac{y_1}{x_1}x$ because $x_1$ may be zero.

Now let's plug $(x_2,y_2)$ in it: $$ x_1 y_2 = y_1 x_2 $$

The radius is the distance from the point to the origin. I'll call it $q$ instead of $x$ to avois confusion. This is given by $$ {x_2}^2 + {y_2}^2 = q^2 $$

Now you have two equations in two unknowns. Substituting the second in the first: $$ {x_1}^2 (q^2-{x_2}^2) = {y_1}^2 {x_2}^2 $$

which is a second degree equation in $x_2$. The corresponding $y_2$ is found with the first equation. It has two solutions, because there are two points in the same line $\bar{OP}$ at the same distance from the origin. If you want, you can discard the solution pointing in the opposite direction.

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Suppose $P$ is at location $(a,b)$, with $a,b$ assumed positive. You want point $P$ to be at $(c,d)$, with $c^2+d^2=x$, but also you want $\frac{d}{c}=\frac{b}{a}$.

We divide both sides by $c^2$ to get $1+\left(\frac{d}{c}\right)^2=\frac{x}{c^2}$, or $$1+\left(\frac{b}{a}\right)^2=\frac{x}{c^2}$$

We can now solve for $c$ via $$c=\sqrt{\frac{x}{1+(b/a)^2}}$$

Having solved for $c$, we may solve for $d$ via $$d=\sqrt{x-c^2}$$

I leave the cases where $a$ and/or $b$ is nonpositive for you to consider as variations.

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