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(B) The given equation of the circle is $x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0$. Comparing this with the standard form $x^2+y^2+2gx+2fy+c=0$,we

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$10$

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(B) The given equation of the circle is $x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0$. Comparing this with the standard form $x^2+y^2+2gx+2fy+c=0$,we get $g=\lambda$ and $f=-\lambda$. The centre is $(-g, -f) = (-\lambda, \lambda)$. Since the circle is in the second quadrant and touches the coordinate axes,the radius $r = |\lambda| = \lambda$ (as $\lambda > 0$ for the second quadrant). The line is $\frac{x}{5}+\frac{y}{12}=1$,which simplifies to $12x+5y-60=0$. The perpendicular distance from the centre $(-\lambda, \lambda)$ to the line $12x+5y-60=0$ must be equal to the radius $\lambda$. $r = \frac{|12(-\lambda)+5(\lambda)-60|}{\sqrt{12^2+5^2}} = \lambda$. $\frac{|-7\lambda-60|}{13} = \lambda$. $|-7\lambda-60| = 13\lambda$. This gives two cases: $1) -7\lambda-60 = 13\lambda$ $\Rightarrow 20\lambda = -60$ $\Rightarrow \lambda = -3$. $2) -7\lambda-60 = -13\lambda$ $\Rightarrow 6\lambda = 60$ $\Rightarrow \lambda = 10$. Since the centre is $(-\lambda, \lambda)$ and it lies in the second quadrant,$\lambda$ must be positive. Therefore,$\lambda = 10$.

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

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