Describing family of circular arcs of varying radii all passing through two points $(0,0)$ and $(1,1)$? I am trying to generate symmetric "ovals" by performing non-linear transformations on ellipses of the same overall dimensions.
"Ovals" are used in designing tabletops and windows but are not defined mathematically (as far as I know.) Laying out an oval with four arcs is the most common way of creating one, but those lack varying curvature and visually look terrible. Ellipses have the required varying curvature but are too "pointed" for a tabletop or window and are not as visually appealing as a well-drawn oval.
I have had some success using a parabolic transformation function $f(x) = abs(x-1)^n + 1$
The procedure I use is to generate a series of points (in Excel) describing an ellipse, and then taking the x-coordinates, normalizing them down to the domain $-1 <= x <= +1$, running them through the above transformation, and then expanding the results back to the original domain.
This procedure expands the central portion of the ellipse while compressing the pointed ends of the ellipse. It works pretty well but that function doesn't have the constant curvature that a circular arc transformation function would have.
So my question is:
I am looking for a different transformation function that has more of a constant curvature than the function described above.
Is is possible to define a function that describes a family of circular arcs of varying radii that all pass through the two points $(0,0)$ and $(1,1)$ such that $r > 0.707$?
 A: There is a one-parameter family of circles through those two points. These circles have equations of the form $$x(x-1)+y(y-1)+\lambda(x-y)=0.\tag1$$ This is basically the sum of the equation of a circle with diameter given by the two points with an arbitrary constant times the equation of the line through the points. Rearranging into a more standard form, we have $$\left(x-\frac12(1-\lambda)\right)^2+\left(y-\frac12(1+\lambda)\right)^2=\frac12(1+\lambda^2),$$ so the center of the circle is at $\frac12(1-\lambda,1+\lambda)$ and its radius is $\sqrt{{1+\lambda^2\over2}}$.  
It might be more convenient to parameterize the family by radius or center position. One can reparameterize by radius $r$ by substituting $\lambda=\pm\sqrt{2r^2-1}$ into either circle equation. The centers of all of the circles lie on the perpendicular bisector of the two fixed points, namely, the line $x+y=1$. A possibly convenient parameterization might be by the (signed) distance $d$ from the from this line, i.e., from the midpoint $(1/2,1/2)$: this puts the center at $\left(1/2+d/\sqrt2,1/2-d/\sqrt2\right)$ and equation (1) becomes $$x(x-1)+y(y-1)-\sqrt2d(x-y)=0.\tag2$$ The circle’s radius is then just the distance of this center point from the origin, namely $\sqrt{d^2+1/2}$.
With the center and radius in hand, it shouldn’t be too hard to work up a parameterization for the arc, if needed.
