The sum of diameters of the circles that touch (i) the parabola 75x² = 64(5y – 3) at the point $\left(\frac{8}{5},\frac{6}{5}\right)$ and (ii) the y-axis, is equal to _______.

${\mathrm{x}}^2=\frac{64.5}{75}\left(\mathrm{y}-\frac{3}{5}\right)$

equation of tangent at $\left(\frac{8}{5},\frac{6}{5}\right)$

$\mathrm{x}\cdot \frac{8}{5}=\frac{64}{15}\left(\frac{\mathrm{y}+\frac{6}{5}}{2}-\frac{3}{5}\right)$

3x – 4y = 0

equation of family of circle is

${\left(\mathrm{x}-\frac{8}{5}\right)}^2+{\left(\mathrm{y}-\frac{6}{5}\right)}^2+\mathrm{\lambda }(3\mathrm{x}-4\mathrm{y})=0$

It touches y axis so f² = c

$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x}\left(3\lambda -\frac{16}{5}\right)+\mathrm{y}\left(-4\lambda -\frac{12}{5}\right)+4=0\\ \frac{{\left(4\lambda +\frac{12}{5}\right)}^2}{4}=4\\ \lambda =\frac{2}{5}\text{ or }\lambda =-\frac{8}{5}\\ \lambda =\frac{2}{5},\mathrm{r}=1\\ \lambda =-\frac{8}{5},\mathrm{r}=4\\ {\mathrm{d}}_1+{\mathrm{d}}_2=10\end{array}$