Find angle of line along a circle's perimeter?

Question

If I know coordinates of point $A$, coordinates of circle center $B$ and $r$ is the radius of the circle, is it possible to calculate the angle of the lines that are passing through point A that are also tangent to the circle?

$A$ is the green point, $B$ is the center of the red circle and I am trying to find out the angle of the blue lines.

Answer 1

Yes! It is possible. Given your set-up, it is a well-known theorem of Euclidean Geometry that tangent segments to a circle from a point outside the circle (in this case, the point $A$) are equal in length. Let us denote the length of such a segment by $b$.

Now, let us denote with $M$ and $N$ the points of tangency of the blue lines in your image (i.e. $M$ will be the point at which one blue line touches the circle and $N$ will be the point at which the other line touches the circle).

By what I said previously, we know that the length of $AM$ will be $b$ and the length of $AN$ will be $b$ as well. Now, since $M$ and $N$ are points of tangency, we know that $\angle M$ and $\angle N$ are right angles. Hence, the triangles $\triangle ABN$ and $\triangle ABM$ are right triangles. Letting $r$ be the length of the radius of the circle you are given, we can use the pythagorean theorem on the triangles $\triangle ABN$ or $\triangle ABM$ to conclude that $b^2 + r^2 = (AB)^2 \implies b = \sqrt{(AB)^2 - r^2}$

Now, we observe that the measure of $\angle A$, denoted $m(\angle A)$, will be the sum of $m(\angle BAM) + m(\angle BAN)$. That is,

$$m(\angle A) = m(\angle BAM) + m(\angle BAN)$$

However, since we know that the lengths of the sides of triangles $\triangle ABN$ and $\triangle ABM$ are the same, we may conclude by the SSS congruence condition that $\triangle ABN \cong \triangle ABM$. Hence, $m(\angle BAM) = m(\angle BAN)$.

Thus, $m(\angle A) = 2m(\angle BAM)$.

However, since $\triangle BAM$ is a right triangle, $m(\angle BAM) = \arctan(\frac{r}{b})$

Thus, $m(\angle A) = 2\arctan(\frac{r}{b})$. Since we know that $b = \sqrt{(AB)^2 - r^2}$, we have:

$$m(\angle A) = 2\arctan(\frac{r}{\sqrt{(AB)^2 - r^2}})$$

Since we know the coordinates of $A$ and $B$, we can simply use the distance formula to find the length $AB$.

Answer 2

Let $P$ be one of the tangency points and $Q$ the other one. Let $\alpha$ be the angle we are searching for. It's well known that $AQ=AP$, then, if we consider the cuadrilateral $APBQ$ we can easily see that $AB$ bisects $\alpha$, then, by looking at the right triangle $APB$ we notice that $\sin\frac{\alpha}{2}=\frac{r}{AB}$, by applying half angle formula we get

$$\pm\sqrt{\frac{1-\cos\alpha}{2}}=\frac{r}{AB}$$ $$\frac{1-\cos\alpha}{2}=\frac{r^2}{(AB)^2}$$ $$\cos\alpha=1-\frac{r^2}{(AB)^2}$$

And since $0<\alpha<\pi$

$$\alpha=\arccos\left(1-\frac{r^2}{(AB)^2}\right)$$

Where $r$ is known and $AB$ is easy to calculate given the coordinates of $A$ and the coordinates of $B$

Answer 3

Yes.

Assume all angles are measured in degrees.

Label the intersection points of the circle and the tangents $T_1$ and $T_2$.

Let $\theta_1=\angle ABT_1$, $\theta_2=\angle ABT_2$. Thus $$\theta_1=\theta_2$$ Therefore the angle in question, $\phi=\angle T_1AT_2$ can be evaluated as follows: $$\phi=2(90-\theta_1)=2(90-\theta_2)$$