Find a circle perpendicular to two other circles.

Question

Here are two circles $C_1 = [ x^2 + y^2 = 1]$ and $C_2 = [ (x-2)^2 + y^2 = 3 ]$. The radii are $1$ and $3$ and the two circles are orthogonal to each other $C_1 \cdot C_2 = 0$. What is the equation of a third circle cutting the other two at right angles? I suspect there are only two solutions which are equivalent by a reflection.

Looking for a geometric solution if possible. Counting the number of solutions, finding the equation of the circles and maybe some insights into the method.

Related: Find a circle orthogonal to two other circles

---

The Inner Product of Two Circles

If $C = [ a(x^2 + y^2) - 2px - 2qy + r = 0]$ the inner product between two such circles would be: $$ [C_1 \cdot C_2] = \frac{1}{2}[2p_1p_2 + 2q_1q_2 - r_1a_2 - r_2a_1] $$

So we are trying to find three circles $C_1, C_2, C_3$ the first two given, such that: $$ [C_1 \cdot C_2] = [C_2 \cdot C_3] = [C_3 \cdot C_1 ]= 0 $$ The inner product can also be written in terms of the radii of the circles and the distance between the centers: $$ [C_1 \cdot C_2 ] = \frac{1}{2} [ R_1^2 + R_2^2 - D^2]$$ and if $C_1 \perp C_2$ we have that this inner product is zero, $[C_1 \cdot C_2 ] = 0$.

Not too many resources on the "algebra of circles", e.g. Geometry a Comprehensive Coruse by Daniel Pedoe.
Descartes theorem deals with mutually tangent circles.

Answer 1

Inversive Geometry

We can find the family of perpendicular circles using Inversive Geometry. If we invert the circles $x^2+y^2=1$ (green) and $(x-2)^2+y^2=3$ (blue) through a circle (gray) whose center is their intersection at $\left(\frac12,\frac{\sqrt3}2\right)$ and that passes through their intersection at $\left(\frac12,-\frac{\sqrt3}2\right)$, the circles are mapped to a pair of (dashed) lines.

---

Perpendicular Circles

Since inversion is conformal, to find a circle perpendicular to both the green and blue solid circles, we simply need to find the inverse of a circle perpendicular to both the green and blue dashed lines. That is, any circle centered at the intersection of those lines.

The red dashed circles are centered at the intersection of the green and blue dashed lines (which are the inverses of the green and blue solid circles), and are perpendicular to both the green and blue dashed lines. The red solid circles are the inverses of the red dashed circles, and are therefore perpendicular to both the green and blue solid circles.

---

Computing The Inverse Circles

Since the inversion center and the centers of the red dashed circles are on the line $x=\frac12$, the centers of the red solid circles will also be on that same line. Given that the radius of the inversion circle is $\sqrt3$ and the radius of the dashed red circle is $r$, we have

1. a point at a distance of $\sqrt3-r$ below the inversion point is mapped to a point at at distance of $\frac3{\sqrt3-r}=\frac{3\left(\sqrt3+r\right)}{3-r^2}$ below the inversion point (negative values are above).

2. a point at a distance of $\sqrt3+r$ below the inversion point is mapped to a point at at distance of $\frac3{\sqrt3+r}=\frac{3\left(\sqrt3-r\right)}{3-r^2}$ below the inversion point (negative values are above).

The radius of the red solid circle is half the distance between these points: $\left|\frac{3r}{3-r^2}\right|$.

The center of the red solid circle is the average of these distances below (or above for negative values) the inversion center: $\frac{\sqrt3}2-\frac{3\sqrt3}{3-r^2}=-\frac{\sqrt3}2\frac{3+r^2}{\left(3-r^2\right)}$

For the red dashed circle centered at $\left(\frac12,-\frac{\sqrt3}2\right)$ with radius $r$, the corresponding red solid circle is centered at $\left(\frac12,-\frac{\sqrt3}2\frac{3+r^2}{3-r^2}\right)$ with radius $\left|\frac{3r}{3-r^2}\right|$. As $r\to\sqrt3$, the red solid circle limits to the $x$-axis.

The equation for this red solid circle is $$ \left(x-\frac12\right)^2+\left(y+\frac{\sqrt3}2\frac{3+r^2}{3-r^2}\right)^2=\left(\frac{3r}{3-r^2}\right)^2 $$

Answer 2

Consider the family of circles with the same two common points,

$$\lambda(x^2+y^2-1)+(1-\lambda)((x-2)^2+y^2-3)=0$$ (these are indeed circles as the $x^2/y^2$ coefficients are equal and there are no $xy$ terms).

Now we eliminate $\lambda$ to get the ODE of the family:

$$\lambda=\frac{(x-2)^2+y^2-3}{2-4x}$$ and by differentiation,

$$((x-2)+yy')(1-2x)-((x-2)^2+y^2-3)(-1)=yy'(1-2x)+y^2-x^2+x-1=0.$$

Now we replace the slope $y'$ by the orthogonal slope and get the differential equation of the orthogonal family

$$y(2x-1)+y'(y^2-x^2+x-1)=0.$$

This is

$$y(2x-1)-y'(x^2-x+1)+y'y^2=y^2\left(\frac{x^2-x+1}{y}+y\right)'=0$$

giving the solutions

$$x^2+y^2-x-cy+1=0,$$

another family of circles, with centers $(\frac12,\frac c2)$ and squared radii $\frac{c^2-3}4$.

---

In fact, these two families are https://en.wikipedia.org/wiki/Circles_of_Apollonius.

Answer 3

Here's a construction using simple high-school geometry:

We start with an arbitrary pair of intersecting circles with centers at $A$ and $C,$ intersecting at $B$ and $D.$ Construct the straight line through $B$ and $D.$ Choose any point on that line that is not on the segment between $B$ and $D.$ For example, consider point $E$ as shown in the figure. Construct a circle on the diameter $AE$. (You can do this by finding the midpoint $F$ of the segment $AE$ and constructing a circle about $F$ through $A$.) Let $G$ be either intersection of the circle about $F$ with the circle about $A.$ Then $\triangle AGE$ is a right triangle with right angle at $G,$ since $G$ is on the circle with diameter on side $AE.$ Hence the circles about $A$ and $E$ are orthogonal.

I leave it as an exercise to prove that the circles about $C$ and $E$ also are orthogonal. A hint: the line $BD$ is the locus of all points $P$ such that $(CP)^2 - (AP)^2 = (CB)^2 - (AB)^2.$

We find then that there are infinitely many solutions. The construction does not work if $E$ is between $B$ and $D$, because then there are no tangents from $E$ to the circles about $A$ and $C$ and hence no orthogonal circles. But as long as $E$ is outside that segment, the construction works.

The construction is a bit trickier if the original circles do not intersect. One way to do it is to find the two external common tangents of the two given circles; let the points of tangency be $P$ and $Q$ for one tangent and $R$ and $S$ for the other tangent. Then the line through the midpoints of segments $PQ$ and $RS$ is the locus of the center of the orthogonal third circle. Of course this construction also works if the original circles intersect.

Answer 4

Length of tangent from circle $C(x,y) = x^2 +y^2 + 2gx + 2fy+ c=0$ from a point $P=(x_i,y_i)$ is given as: $C(P) = C(x_i,y_i)$

Now, if we have two circles $C_1$ and $C_2$, then the radical axis is given as:

$$C_1(x,y) = C_2(x,y)$$

Or,

$$ 2(g_1 - g_2) x + 2(f_1 - f_2)y + c_1 - c_2 = 0 \tag{1}$$

I.e: points for which length of tangent from each circle is equal.

Now, for any point on this line, we can draw a circle which intersects the two original circles orthogonally (Picture from wikipedia). We can write a point for $(1)$ parametrically as $$(t, \frac{c_2 -c_1}{2(f_1 - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)$$, now a circle cantered at a point with radius $r$ as given as:

$$ (x-t)^2 + (y-\frac{c_2 -c_1}{2(f_1 - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)^2 = r^2$$

But what is the radius from the definition of radical axis? it is the length of tangents!

Hence,

$$ r(t)=\sqrt{C_1(t,\frac{c_2 -c_1}{2(f_1 - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t) }= \sqrt{C_2(t,\frac{c_2 -c_1}{2(f_1 - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)} $$

Which leads to:

$$ (x-t)^2 + (y-\frac{c_2 -c_1}{2(f_1 - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)^2= C_1(t,\frac{c_2 -c_1}{2(f_1 - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t) =C_2(t,\frac{c_2 -c_1}{2(f_1 - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)$$