# family of circles in bipolar coordinate system

I don't get the idea how the equation for this family of curve is $\displaystyle y^2 + (x - a \coth v)^2 = \frac{a^2}{\sinh ^2v}$ from this article on Wikipedia. Suppose, the equation is $\displaystyle y^2 + (x-h)^2 = r^2$, how do I parametrize $h$ and $r$? What points should I check to get in this form?

The circle seems to have radius $0$ when it passes through $(\pm a, 0)$ and infinite radius at $(0, 0)$. What would be equation of this family of curve in Cartesian coordinate?

• So, what happened after trying the Cartesian-bipolar conversion formula on $\tau=c$? – J. M. is a poor mathematician May 29 '13 at 8:38
• @J.M. don't know never tried that (looks complicated), i believe it should have simple way of doing, for the other one seems obvious though as they both passes though $(-a,0)$ and $(a, 0)$. which reduces to $a^2 + a^2 k^2 = r^2$. For this particular, I can't find that it would have this form. – hasExams May 29 '13 at 8:42
• But, can you at least leverage the usual Pythagorean formula to eliminate $\sigma$? – J. M. is a poor mathematician May 29 '13 at 8:46
• @J.M. probably that definition follows from this curve relation. It's exactly opposite on my book. First the two curves and then transformation equations. – hasExams May 29 '13 at 8:48
• @J.M. looks like you are right!! – hasExams May 29 '13 at 8:49