If the curves x^2 - 6x + y^2 + 8 = 0 and x^2 | 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 is $C_1(3, 0)$ an

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 is $C_1(3, 0)$ and radius $r_1 = \sqrt{3^2 + 0^2 - 8} = \sqrt{9 - 8} = 1$.For $S_2$,the center is $C_2(0, 4)$ and radius $r_2 = \sqrt{0^2 + 4^2 - (16 - k)} = \sqrt{16 - 16 + k} = \sqrt{k}$.The distance between the centers is $d = \sqrt{(3 - 0)^2 + (0 - 4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.Two circles touch each other if $d = r_1 + r_2$ (externally) or $d = |r_1 - r_2|$ (internally).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 of $k$,the largest value 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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