# Finding a point on a circle that makes a specific angle of incidence with a given point inside it?

Suppose we have a circle of radius r centered at the origin, and we are given a point C$(x,y)$ inside the circle.

Let $\theta_i$ be the angle of incidence of the line drawn from C. If we know $\theta_i$ (say, 23º) and C, how can we find the coordinates of P$(x,y)$, the point on the circle where the angle of incidence from C is $\theta_i$?

Edit 1: Oops, forgot to draw P in the circle. It's the point where the line meets the circle.

• I haven't done the calculations so I'm posting this as a comment. This is the strategy: Let the point be P(x,y). Make two linear equations for the two lines OX and OP, then use the formula $$\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ (where m1 and m2 are the slopes of the lines) and solve for theta. Oct 6, 2015 at 16:40
• @Kartik The problem is that your unknowns are in the $m$s. Oct 6, 2015 at 16:52
• Just to clarify, you know only the coordinates of $C$, and the angle of incidence $\theta_i$, and you want to know the coordinates of $P$, the point where the "reflected" ray strikes the circle again (in this diagram, on the right side of the circle, just above the $x$-axis)? Oct 6, 2015 at 17:03
• No, I know $C$, and I want to find the point $P$ on the circle which has an angle of incidence of the given $θ_i$. The reflected ray is actually extraneous; apologies for the confusion. Oct 7, 2015 at 0:55

To find the points $P$ on a given circle $c$ of center $O$ and radius $R$, such that $\angle CPO=\alpha$ given, join $CO$ and from its midpoint $M$ draw the perpendicular bisector $r$. Let $A$ be a point on $r$ such that $\angle CAM=\alpha$ (that is $AM=CM/\tan\alpha$), then draw the circle $a$ having center at $A$ and passing through $O$ and $C$. The intersections between $a$ and the given circle $c$ are the required points, as well as the symmetric of those points with respect to line $OC$. In fact, if $P$ is one of such intersections, then $P$ lies on $a$ and $\angle CPO=(1/2)\angle CAO=\alpha$. Notice that we have no solutions for $OC<R\sin\alpha$ ($a$ contained in $c$), two solutions for $OC=R\sin\alpha$ ($a$ tangent to $c$) and four solutions for $OC>R\sin\alpha$ ($a$ secant to $c$).

If $x_C$, $y_C$ are the coordinates of $C$, then the coordinates of $A$ (or $A'$) are given by $x_A={1\over2}(x_C\mp y_C/\tan\alpha)$, $y_A={1\over2}(y_C\pm x_C/\tan\alpha)$, and it is straightforward to get the coordinates of the intersection points: $$x_P={2R\sin^2\alpha\over t^2} \left( Rx_A\pm y_A\sqrt{{t^2\over\sin^2\alpha}-R^2} \right),\ y_P={2R\sin^2\alpha\over t^2} \left( Ry_A\mp x_A\sqrt{{t^2\over\sin^2\alpha}-R^2} \right),$$ where $t=OC=\sqrt{x_C^2+y_C^2}$.

Edit from Blue. Here's a picture:

• @Blue Thanks a lot, the picture is perfect now! Oct 7, 2015 at 11:14
• @Aretino Very interesting diagram. Is it possible to construct a formula from given values (e.g. $O$, $C$, r, $\theta_i$) to find one of the coordinates of $P$? Oct 7, 2015 at 12:02
• I've just added some formulas: please check them before use. Oct 7, 2015 at 14:19
• Just to clarify: $R$ is the radius of the yellow circle, right? Oct 8, 2015 at 13:17
• And of course I hope the sense of the various $\pm$ is clear enough: one has two possible positions for point $A$, and for each choice of $A$ one has then two possible intersection points (if they exist, of course). Oct 8, 2015 at 16:20

Assume that the center of the circle is the origin. We have $P\cdot(P-C) = |P| |P-C| \cos\theta$, but $|P|=1$, so $P\cdot (P-C) = P\cdot P - P\cdot C$ (by linearity) $= 1-P\cdot C$ and this equation becomes $1-P\cdot C = |P-C|\cos\theta$; then squaring both sides and using $|v|^2=v\cdot v$, we get $(1-P\cdot C)^2 = \cos^2\theta (P-C)\cdot (P-C)$. We can use linearity to expand the RHS again, getting $(1-P\cdot C)^2=\cos^2\theta(P\cdot P-2P\cdot C+C\cdot C) = \cos^2\theta (1+|C|^2-2P\cdot C)$. This is a quadratic equation in $P\cdot C$, which you can solve to get that value (or values - I see no reason that the given point $P$ should be unique).

Finally, given the equations $P\cdot C=q$ (where $q$ is a calculated value from solving the quadratic above) and $P\cdot P=1$, there are presumably several ways to solve for $P$ but the simplest (to me) is to just set $P=(x,y)$ and note that $P\cdot C=q$ is a linear equation in $x$ and $y$, so we can use it to substitute one of $x,y$ for the other in $P\cdot P=1$ (which is just the equation $x^2+y^2=1$ of the circle), getting the final values of $x,y$.

• $P$ is not unique in general, depending on whether $C$ is closer to the point of incidence than is the center of the circle, or further. Or, possibly, it might not exist, if $C$ is to close to the center of the circle. Oct 6, 2015 at 17:08
• @BrianTung Yep, that makes sense to me. The conditions here, expanded out - I didn't actually plug in the full quadratic for $P\cdot C$ - should reflect those constraints. Oct 6, 2015 at 17:23

Let $Q$ be the point of incidence. We can solve the triangle OCQ with sides $r$ and $|OC|$ and the angle $\theta$ by the sine law and find the angle $\angle{COP}$. $$\sin(\angle OCQ)=\frac r{|OC|}\sin(\theta)\\ \angle COQ=\pi-\angle OCQ-\theta.$$

This gives the polar angle of $Q$ (knowing that of $C$). The requested point $P$ has a polar angle equal to that of $Q$ minus the angle at the apex of an isosceles triangle, $\pi-2\theta$.

• Fom the figure below we can calculate:

$\cos(D)=\frac{a}{d}$.....(1)

$\cos(D+\theta)=\frac{a}{r}$ ..... (2)

(2) in (1) -> $\cos(D)=\frac{r\cos(D+\theta)}{d}$

$d=\frac{r\cos(D+\theta)}{\cos(D)}$ ........... (3)

1-the big triangle:

$\frac{r}{\sin(e)}=\frac{B}{\sin\theta}$

$\sin(e)=\frac{r\sin\theta}{B}$.....(4)

$\frac{C}{sin(Y+y)}=\frac{C}{\sin(2\pi-e-\theta)}=\frac{C}{\sin(-e-\theta)}=-\frac{C}{\sin(e)\cos\theta+\cos(e)\sin\theta}=\frac{B}{\sin\theta}$

$C=-B\sin(e)cot\theta-B\cos(e)$....(5)

2- from right triangle:

$\frac{A}{\sin(e)}=\frac{B}{\sin(\pi-t)}$

$A\sin(t)=B\sin(e)$,

from (4): $A\sin(t)=B\frac{r\sin\theta}{B}=r\sin\theta$....(6)

$\frac{A}{\sin(e)}=\frac{C-d}{\sin(Y)}$

$\frac{A\sin(Y)}{C-d}=\sin(e)$

From (5), $\frac{A\sin(Y}{C-d}=\frac{A\sin(Y)}{-B\sin(e)\cot\theta-B\cos(e)-d}=\sin(e)$,

using (4),

$A\frac{sin(-Y)}{B\frac{r\sin\theta}{B}\cot\theta+B\sqrt{1-\frac{r^2\sin^2\theta}{B^2}}+d}=\frac{r\sin\theta}{B}$ ...... (7)

3- From left triangle:

$\frac{A}{\sin(\theta)}=\frac{d}{\sin(y)}$

$\frac{A*\sin(y)}{d}=\sin\theta$ ..... (8)

• Knowing that $\sin(y)=\frac{a}{r}=\cos(D+\theta)$ ....(9)

(3),(9) in (8) -> $\frac{A*\cos(D+\theta)}{\frac{r\cos(D+\theta)}{\cos(D)}}=\sin\theta$

$\frac{A*\cos(D)}{r}=\sin\theta$ .... (10)

(3) in (7) ,

$A\frac{sin(-Y)}{B\frac{r\sin\theta}{B}\cot\theta+B\sqrt{1-\frac{r^2\sin^2\theta}{B^2}}+\frac{r\cos(D+\theta)}{\cos(D)}}=\frac{r\sin\theta}{B}$ ...... (11)

(10) in (11)

$\frac{r\sin\theta}{\cos(D)} \frac{sin(-Y)}{B\frac{r\sin\theta}{B}\cot\theta+B\sqrt{1-\frac{r^2\sin^2\theta}{B^2}}+\frac{r\cos(D+\theta)}{\cos(D)}}=\frac{r\sin\theta}{B}$

${B}\frac{sin(-Y)}{{\cos(D)}{r\cos\theta}+{\cos(D)}\sqrt{B^2-{r^2\sin^2\theta}}+{r\cos(D+\theta)}}=1$

$\cos^2 D + \frac{{\left(\cos D\, \left(2\, r\, \mathrm{\cos}\!\left(\mathrm{\theta}\right) + \sqrt{B^2 - r^2\, {\mathrm{\sin}\!\left(\mathrm{\theta}\right)}^2}\right) - B\, \mathrm{\sin}\!\left(- Y\right)\right)}^2}{r^2\, {\mathrm{\sin}\!\left(\mathrm{\theta}\right)}^2} - 1=0$...(12)

with:

$x=cos(D)$

$a1=2*r*\cos(\theta)+\sqrt{B^2-(r^2*\sin(\theta)^2)}$

$a2=(B)*\sin(-Y)$

$a3=(r*\sin(\theta))^2$

solutions:

$x_1=\frac{\mathrm{a1}\, \mathrm{a2} + \mathrm{a3}\, \sqrt{\frac{{\mathrm{a1}}^2 - {\mathrm{a2}}^2 + \mathrm{a3}}{\mathrm{a3}}}}{{\mathrm{a1}}^2 + \mathrm{a3}}$

$x_2=\frac{\mathrm{a1}\, \mathrm{a2} - \mathrm{a3}\, \sqrt{\frac{{\mathrm{a1}}^2 - {\mathrm{a2}}^2 + \mathrm{a3}}{\mathrm{a3}}}}{{\mathrm{a1}}^2 + \mathrm{a3}}$

• When typesetting try to use \sin instead of sin because it is easier on the eyes. Like $\sin(a)\cos(b)$ vs $sin(a) cos(b)$. Oct 7, 2015 at 0:54