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Equations $C_1 = 0, C_2 = 0$ represent two circles in the plane.

$C_1 - C_2 = 0$ represents a radical axis between the circles of equal

tangents/power, and $C_1 + C_2 = 0$ represents the another circle placed in between the circles.

What geometric property is averaged in the summed up circle from the component circles?

Does it still hold good if $C$ represents a conic instead of a circle? If not, how is the generalized situation?

Does $\lambda C_1 + \mu C_2 = 0 $ constitute a homotopy?

EDIT1:

In an attempt to make sense generalizing circles to conics:

How or along what line is the center of union ellipse $ \lambda C_1 + \mu C_2 = 0 $ located connecting the centers of the two ellipses $C_1$ and $C_2$ ?

What geometrical objects from the union ellipse touch ellipses $ C_1 , C_2 $ in an equitable manner?

Narasimham
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  • I'm unsure what you meant by “circles of equal tangents/power”. Two circles always have at least two tangents in common, and the power of a circle sounds like something else entirely. I also don't understand your last question about the homotopy. Isn't homotopy always a relation between two objects? How can a single geometrical object be a homotopy? I may be able to extend my answer if you can clarify these points. – MvG Nov 10 '16 at 12:33
  • The radical axis is locus of equal tangent length squared, defined as power for either of two circles or a set of them. $ x^2+y^2+2 f x+2 g y+ P=0 $ has power $P$ = square of tangent lengths from origin. It is $>0$ for points on radical axis outside the circle,$ <0 $ when inside. Regarding homotopy, I included all circles on common radical axis with same power in generalization. Should I improve upon the posted question ? – Narasimham Nov 10 '16 at 16:29
  • Ah, so “equal tangents/power” was a description of point son the radical axis, not a property the two circles should have. Makes more sense now. The homotopy I still don't get. Changing the $C_1+C_2 = 0$ to a $C_2+C_2 = \text{const.}$ will merely change the radius of the circle. As these are concentric their pairwise radical axis is the line at infinity. And the radical axis of one such circle with $C_1$ or $C_2$ can be any parallel to the radical axis of $C_1$ and $C_2$. So I don't see “all circles on common radical axis” here. If you can, please edit question to improve. – MvG Nov 10 '16 at 16:42
  • Yes. Changing $C_1+C_2$ from 0 to non-zero constant in homotopy generates a set of circles of differing radii passing through two concurrent points, real or complex.They are not concentric. – Narasimham Nov 10 '16 at 18:25
  • Yes they are concentric. If you take $\tfrac12(C_1+C_2)$ as $(x^2+y^2)+ax+by+c$ then changing the radius only will only change $c$, i.e. the constant term. So if you have a constant on the right, that's equivalent to reducing $c$, so you only change the radius. The conic analogon here is a pencil of conics sharing two points and the tangent directions in those points, except that the points in this case are the ideal circle points $[1:\pm i:0]$. The tangents in these points meet in the center of the circle. – MvG Nov 10 '16 at 18:44
  • Apologies, went off course because $C_1+C_2=$ constant circles are not relevant to the question. Referring to homotopy as $\lambda C_1+ \mu C_2= 0 $ generating a set of circles of differing radii, all passing through two concurrent points real or complex, on radical axis. – Narasimham Nov 11 '16 at 18:16

1 Answers1

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Circles

Your post seems to assume that your equations $C_i$ are of the form $(x^2+y^2)+ax+by+c$ so that the parentheses actually cancel out in the difference to yield the radical axis. In this case, $(-\frac a2,-\frac b2)$ is the center of the circle, as can be seen if you complete the square. So if you sum up two terms like this, you might want to divide by two afterwards so that the result is again of the same form. So you'd average the centers. Furthermore, any point which satisfies $C_1=0$ and satisfies $C_2=0$ also satisfies $C_1+C_2=0$ (and indeed satisfies $\lambda C_1+\mu C_2=0$ for arbitrary $\lambda,\mu$).

So $C_1+C_2=0$ is the circle around the midpoint between the centers of $C_1$ and $C_2$, with a radius such that it passes through the points of intersection of $C_1$ and $C_2$. Which may be complex, leading to a circle of complex radius, as opposed to the radical axis which stays real as the imaginary contributions cancel out.

Conics

For conics, it doesn't make much sense to require a $1$ as the coefficient of the term $(x^2+y^2)$ since $x^2$ and $y^2$ might have different coefficients. So I'd say a generic conic is given by $ax^2+by^2+cxy+dx+ey+f=0$, with at least one non-zero coefficient. In that case, the sum or difference of two conics depends entirely on how you scale your equations. But no matter how you cale them, any linear combination of two given conics will lead to conics which share the same four points of intersection (some of which might be complex).

In general, the pencil of conics $\lambda C_1+\mu C_2=0$ will contain three degenerate cases: situations where the conic turns into a pair of lines. These pairs of lines correspond to the ways how you can take the four points of intersection, group them in pairs and connect the two points from each pair to form one line of the degenerate conic. Actually this is often used in the reverse direction: find the degenerate conics, decompose them into lines and intersect these lines to obtain the points of intersection. See this post of mine for details.

Often, it makes sense to choose $C_1$ and $C_2$ as such degenerate conics through four given points, in such a way that $C_1+C_2$ yields the third degenerate conic. Then these establish a (somewhat canonical) projective basis for the pencil of conics: any conic $\lambda C_1+\mu C_2=0$ from the pencil could be described by a homogeneous coordinate vector $[\lambda:\mu]$ with $[1:0]$, $[0:1]$ and $[1:1]$ used as the canonical basis. This structure is isomorphic to the projective line.

Centers of conics

As your edit extends the question to ask about the center of the conic $\lambda C_1+\mu C_2=0$, here is how I'd think about this. The conic $ax^2+by^2+cxy+dx+ey+f=0$ can also be written as a symmetric matrix:

$$(x,y,1)\cdot\begin{pmatrix}2a&c&d\\c&2b&e\\d&e&2f\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0$$

The center of symmetry of a conic is the pole of the line at infinity, which in turn is the intersection of the polar lines of two points at infinity. The polar line of a point is this matrix times the homogeneous coordinates of the point. So the first column of the matrix is the polar of $[1:0:0]$ (the point at infinity in $x$ direction), and the second column is the polar of $[0:1:0]$. Their cross product is the center: $[ce-2bd:cd-2ae:4ab-c^2]$ in homogeneous coordinates.

Now the center of $\lambda C_1+\mu C_2=0$ is the center you get from the corresponding linear combination of these matrices. Which means you're essentially looking at something like $(\lambda v_1+\mu v_2)\times(\lambda v_3+\mu v_4)$, with the $v_i$ being the appropriate columns of the appropriate matrices. The result is not a simple linear combination like $\lambda(v_1\times v_3)+\mu(v_2\times v_4)$, as one might naïvely expect. Instead, the resulting locus is again a conic. I know that's the case because there is a method to construct the family of tangents to a conic by connecting corresponding entries in two sets of points, if this correspondence is a perspectivity. Think string art and perhaps see Theorem 10.2 in Perspectives on Projective Geometry. This is one way to geometrically interpret the formula above: $(\lambda v_1+\mu v_2)$ is one set of points, $(\lambda v_3+\mu v_4)$ the other, and the cross product means you connect them. At least in theory I could compute a formula for said conic, perhaps based on the coordinates of the four points of intersection. But I guess I'll leave that as a different question.

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