When two soap bubbles of radii a and b $\left(\text{b > a}\right)$ coalesce, the radius of curvature of common surface is:

$\begin{array}{l}\text{ If radius of interface is }r\text{'}\text{ and radii of each bubble is }{r}_1 and r_2, then\text{ equating pressure on either side of common boundary gives}\\ \left(P_0+\frac{4T}{r_1}\right)-\left(P_0+\frac{4T}{r_2}\right)=\frac{4T}{r\text{'}}\\ \therefore \frac{1}{r\text{'}}=\frac{1}{r_1}-\frac{1}{r_2}\\ \Rightarrow \frac{1}{r\text{'}}=\frac{r_2-r_1}{r_1{r}_2};\text{ here }{r}_1=a, r_2=b\\ \Rightarrow \frac{1}{r\text{'}}=\frac{b-a}{ab}\\ \Rightarrow r\text{'}=\frac{ab}{b-a}\end{array}$