If the curves x^{2}-6x+y^{2}+8=0 and x^{2}-8y | vedclass

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(B) The given equations of the circles are $S_{1}: x^{2}+y^{2}-6x+8=0$ and $S_{2}: x^{2}+y^{2}-8y+16-k=0$.For $S_{1}$,the center $C_{1} = (3, 0)$ and

Vedclass · Accepted Answer

$36$

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(B) The given equations of the circles are $S_{1}: x^{2}+y^{2}-6x+8=0$ and $S_{2}: x^{2}+y^{2}-8y+16-k=0$.For $S_{1}$,the center $C_{1} = (3, 0)$ and radius $r_{1} = \sqrt{3^{2}+0^{2}-8} = \sqrt{1} = 1$.For $S_{2}$,the center $C_{2} = (0, 4)$ and radius $r_{2} = \sqrt{0^{2}+4^{2}-(16-k)} = \sqrt{k}$.The distance between the centers is $d = \sqrt{(3-0)^{2}+(0-4)^{2}} = \sqrt{9+16} = 5$.Since the circles touch each other,the distance between centers must be equal to the sum or difference of the radii: $d = |r_{1} \pm r_{2}|$.Case $1$: $d = r_{1}+r_{2}$ $\Rightarrow 5 = 1+\sqrt{k}$ $\Rightarrow \sqrt{k} = 4$ $\Rightarrow k = 16$.Case $2$: $d = |r_{1}-r_{2}| \Rightarrow 5 = |1-\sqrt{k}|$.This implies $1-\sqrt{k} = 5$ or $1-\sqrt{k} = -5$.$1-\sqrt{k} = 5 \Rightarrow \sqrt{k} = -4$ (not possible as $k>0$).$1-\sqrt{k} = -5$ $\Rightarrow \sqrt{k} = 6$ $\Rightarrow k = 36$.Comparing the values,the largest value of $k$ is $36$.

If the curves $x^{2}-6x+y^{2}+8=0$ and $x^{2}-8y+y^{2}+16-k=0$ $(k>0)$ touch each other at a point,then the largest value of $k$ is

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