If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also the line $\frac{x}{5}+\frac{y}{12}=1\text{ is }{x}^2+y^2+2\lambda x-2\lambda y+{\lambda }^2=0\text{, then }\lambda =$

Let r be the radius of the circle. 

Circle touches coordinate axes and centre lies in second quadrant + Centre C = (- r, r)

circle touch $\frac{x}{5}+\frac{y}{12}=1\Rightarrow \left|\frac{12(-r)+5\left(r\right)-60}{\sqrt{25+144}}\right|=r\Rightarrow \left|\frac{-7r-60}{13}\right|=r$

$\Rightarrow 7r+60=\pm 13r\Rightarrow r=10$

Equation of the circle is $(x+10{)}^2+(y-10{)}^2={10}^2\Rightarrow x^2+y^2+20x-20y+100=0\Rightarrow \lambda =10$