Find the equation of a system of coaxial circles of which the points $(\pm k,0)$ are the limiting points.

Question

Find the equation of a system of coaxial circles of which the points $(\pm k,0)$ are the limiting points.

The definition of some terms relevant to the question in 2D geometry is as follows:

The coaxial circles is the set of all circles such that any two of the circles have the same radical axis.

The radical axis of two circles is the locus of the points which have the same powers with respect to them.(The power of a point with respect to a circle is the length of the tangents drawn from the point to the circle)

My solution goes like this:

The equation of the radical axis which is $y-0=0(x-k)$ and thus, $y=0.$ Now, we know that the limiting points and centres in a coaxial system of circles are collinear. Given, $(\pm k,0)$ are the limiting points, we can say, the centre of any circle in the coaxial system lies on the $x-$ axis. Thus, the equation of the point circles corresponding to limiting points $(\pm k,0)$ are $(x\pm k)^2+y^2=0.$ Thus, the equation of the system of coaxial circles is $$(x-k)^2+y^2+\lambda y=0.$$

Is the above solution correct? If not, where is it going wrong?

Answer 1

Your equation $(x-k)^2+y^2+\lambda y=0$ is not correct: the centers of these circles are not on the $x$-axis.

Following https://mathworld.wolfram.com/CoaxalCircles.html, the equation of a system of coaxial circles with limiting points $(±k,0)$ is: $$x^2+y^2+2\lambda x+k^2=0$$ or equivalently $$(x+\lambda)^2+y^2+k^2-\lambda^2=0.$$

Answer 2

Since $(\pm k, 0)$ are limiting points of the system, these two point circles $(x\pm k)^2+y^2=0\tag*{}$ belong to the system.

Label their LHS as $S_1(x,y)$ and $S_2(x,y)$ then the equation of the system is $S_1+\lambda(S_1-S_2)=0$.

$$x^2+y^2+(2k+4\lambda k)x+k^2=0$$

Since $k\neq 0$, you might want to replace $2k+4\lambda k$ with $\lambda$ (because it's just a parameter).

$$\boxed{x^2+y^2+\lambda x+k^2=0}$$

Visual graph for demonstration.