Let $R_1$ and $R_2$ be the radii of the circles $C_1$ and $C_2$ touching the ellipse $3x^2+4y^2=7$ at the point $P\left(1,1\right)$ and also touching $x-$ axis ; $\left(R_1>R_2\right)$ , Let the point $Q\left(a,b\right)$ on the circle $C_2$ which is farthest form the circle $C_1$, then the value of $3a+9b$ is

Required circles C₁ and C₂ are given by 
$\therefore$(x - 1)² +(y - 1)² + k(3x + 4y - 7) = 0
$\therefore$ It touches x axis 
$\therefore$x2 + (3k - 2)x + (2 - 7k)h = 0 has equal roots 
$\therefore$K = -2, 2/9
C₁ = x² + y² - 8x - 10y + 16 = 0
${\mathrm{C}}_2={\mathrm{x}}^2+{\mathrm{y}}^2-\frac{4}{3}\times \frac{10}{8}\mathrm{y}+\frac{4}{9}=0$
Slope of $\mathrm{AB}=\frac{4}{3},\mathrm{Q}=\left(\frac{6}{3}-\frac{5}{9}\times \frac{3}{5}-\frac{5}{9}\times \frac{4}{5}\right) \mathrm{i}.\mathrm{e}. \left(\frac{1}{3},\frac{1}{9}\right)$