Locus of a Point at which two circles subtend equal angles

Question

The locus of a point at which two given unequal circles subtend equal angles is? Ans- Circle My work: I assumed two circles with different general equations and the required point be P(h,k). Then i drew a pair of tangent to both circles and then equated the angle between both the pair of tangents equal. But I don't know how to eliminate the variables assumed in the circle equations to get the locus. Is my approach right, what further should I do or some other way to do it.

Answer 1

I'm interpreting your question as follows: Find the locus of all $P$ such that if tangents are drawn from $P$ to circle $A$, then the angle between those tangents will be the same as the angle between the tangents drawn from point $P$ to circle $B$.

If I am interpreting your question correctly, you're correct in assuming that the locus is a circle. I'm going to use a geometric proof to show you why.

Call the center of one circle $A$ and the other $B$. Call the radius of the first circle $r_a$, and the radius of the second $r_b$. W.L.O.G. $r_a<r_b$. Mark the intersection of the internal tangents of the circles $I$. Mark the intersection of their external tangents $E$. Call the midpoint of $I$ and $E$ point $K$. Call the circle centered at $K$ with diameter $IE$ circle $O$. Then, the locus of points in your problem is circle $O$.

Proof:

Claim: If $P$ is in the locus, then it is on circle $O$.

Justification: $P$ is on the locus if and only if $\displaystyle \frac{AP}{BP}=\frac{r_a}{r_b}$(can you figure out why?)

Using law of sines, you can also figure out that $\displaystyle \frac{AP}{BP}=\frac{AI}{BI}\cdot\frac{\sin\left(\angle IPB\right)}{\sin\left(\angle IPA\right)}$

Since $\displaystyle \frac{AI}{BI}=\frac{r_a}{b_a}$, we have that $\displaystyle \frac{\sin\left(\angle IPB\right)}{\sin\left(\angle IPA\right)}=1$

Because $\angle IPA + \angle IPB <\pi$, we have $\angle IPA = \angle IPB$

Similarly, using law of sines, you can prove that $\displaystyle \frac{AP}{BP}=\frac{AE}{BE}\cdot\frac{\sin\left(\angle EPB\right)}{\sin\left(\angle EPA\right)}$.

Since $\displaystyle \frac{AE}{BE}=\frac{r_a}{r_b}$, we have that $\displaystyle \frac{\sin\left(\angle EPB\right)}{\sin\left(\angle EPA\right)}=1$

Because $r_a<r_b$, we have that $\angle EPA<\angle EPB$. Ergo, $\angle EPA=\pi-\angle EPB$.

Using what we know, let's try to calculate $\angle IPE$:

\begin{align} \angle IPE &= \angle IPA+\angle EPA\\ &=\angle IPA+\pi-\angle EPB\\ &=\angle IPA+\pi-\left(\angle EPA+\angle BPA\right)\\ &=\angle IPA+\pi-\left(\angle EPA+2\angle IPA\right)\\ &=\pi-\left(\angle EPA + \angle IPA\right)\\ &=\pi-\angle IPE \end{align}

Ergo $\angle IPE=\displaystyle \frac{\pi}{2}$. This implies that $P$ is on the circle with diameter $IE$. In other words, $P$ is on circle $O$.

Claim: If $P$ is on circle $O$, then $P$ is in the locus.

Justification: Using law of sines, you can derive the following equations: $$\frac{\sin\left(\angle IPA\right)}{\sin\left(\angle IPB\right)}=\frac{AI}{BI}\cdot\frac{BP}{AP}=\frac{r_a}{r_b}\cdot\frac{BP}{AP}$$ $$\frac{\sin\left(\angle EPA\right)}{\sin\left(\angle EPB\right)}=\frac{AE}{BE}\cdot\frac{BP}{AP}=\frac{r_a}{r_b}\cdot\frac{BP}{AP}$$

Ergo, we have the following: $$\frac{\sin\left(\angle IPA\right)}{\sin\left(\angle IPB\right)}=\frac{\sin\left(\angle EPA\right)}{\sin\left(\angle EPB\right)}$$ We can rearrange this equation like this: $$\frac{\sin\left(\angle IPA\right)}{\sin\left(\angle EPA\right)}=\frac{\sin\left(\angle IPB\right)}{\sin\left(\angle EPB\right)}$$

We also know that $\displaystyle \angle IPE = \angle EPA + \angle IPA = \angle EPB-\angle IPB =\frac{\pi}{2}$.

So, we can make substitutions to get the following: $$\frac{\sin\left(\angle IPA\right)}{\sin\left( \frac{\pi}{2}-\angle IPA\right)}=\frac{\sin\left(\angle IPB\right)}{\sin\left(\frac{\pi}{2}+\angle IPB\right)}$$

We can simplify this expression like so: $$\frac{\sin\left(\angle IPA\right)}{\cos\left( \angle IPA\right)}=\frac{\sin\left(\angle IPB\right)}{\cos\left(\angle IPB\right)}$$

We can again simplify to get the following: $$\tan\left(\angle IPA\right)=\tan\left(\angle IPB\right)$$

Because $\angle IPA + \angle IPB < \pi$, we have $\angle IPA = \angle IPB$.

Again, we use the the previous equation: $$\frac{AP}{BP}=\frac{r_a}{r_b}\cdot\frac{\sin\left(\angle IPB\right)}{\sin\left(\angle IPA\right)}$$

We can simplify this using $\angle IPA = \angle IPB$: $$\frac{AP}{BP}=\frac{r_a}{r_b}\cdot 1=\frac{r_a}{r_b}$$

This implies that $P$ is in the locus.

Since we proved that the circle is a subset of the locus, and the locus is a subset of the circle, we have that the locus and the circle are one and the same.

Answer 2

I would proceed in this way.

Consider as in the figure, a circle of radius $r$. The points whose tangents to the circle make an angle $2\alpha$ are those lying on a concentric circle of radius $r/sin \alpha$.

Thus given two circles of radii $r$ and $R$, separated by a distance $d$, the points whose tangents make an angle $2\alpha$ with each circle will be those at distance $r/sin\alpha$ and $R/sin\alpha$ from the respective centers. That is $$ OP = r\sin \alpha \quad CP = R\sin \alpha $$ and eliminating $\alpha$ $$ \begin{gathered} \frac{{OP}} {{CP}} = \frac{r} {R} \hfill \\ R^{\,2} \left( {x^{\,2} + y^{\,2} } \right) = r^{\,2} \left( {\left( {x - d} \right)^{\,2} + y^{\,2} } \right) \hfill \\ \left( {R^{\,2} - r^{\,2} } \right)x^{\,2} + 2\,r^{\,2} d\;x + \left( {R^{\,2} - r^{\,2} } \right)y^{\,2} = r^{\,2} d^{\,2} \hfill \\ \end{gathered} $$ which gives: $$ \left\{ {\begin{array}{*{20}c} {\;x = d/2} & {R = r} \\ {\left( {x + \,\frac{{r^{\,2} }} {{\left( {R^{\,2} - r^{\,2} } \right)}}d} \right)^{\,2} \; + y^{\,2} = \frac{{R^{\,2} \,r^{\,2} }} {{\left( {R^{\,2} - r^{\,2} } \right)^{\,2} }}d^{\,2} } & {R \ne r} \\ \end{array} } \right. $$

Answer 3

Here I show the way you can construct the locus without calculating the radius of locus; see figure below:

:

!- Draw common tangent SE of big circle $circle(J, S)$ and $Circle(C, E)$ and it's reflect about SE( segment TV), they intersect at point W which a point with required property.

2- Draw third common tangent NQ, it intersect the line JC connecting the centers of two circles at R, this is another point with the same property.

3- Connect R to W.

4 Draw perpendicular bisector of RW it intersects the extension of JC at point M. his is center of locus.

5- Draw a circle passing point W center on M., this would be the locus.