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(B) The equation of the circle is $x^2+y^2-16x-4y=0$. The centre of the circle is $(8, 2)$.Let the variable line passing through $(8, 2)$ have slope $

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$18$

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(B) The equation of the circle is $x^2+y^2-16x-4y=0$. The centre of the circle is $(8, 2)$.Let the variable line passing through $(8, 2)$ have slope $m$. The equation of the line is $(y-2) = m(x-8)$.Since the line meets the positive coordinate axes at $A$ and $B$,the line must have a negative slope,so let $m = -k$ where $k > 0$.The $x$-intercept $OA$ is found by setting $y=0$: $-2 = m(x-8) \Rightarrow x-8 = -2/m \Rightarrow OA = 8 - 2/m$.Since $m  0)$,then $OA = 8 + 2/k$.The $y$-intercept $OB$ is found by setting $x=0$: $(y-2) = m(-8) \Rightarrow y = 2 - 8m = 2 + 8k$.We want to minimize $f(k) = OA + OB = 8 + 2/k + 2 + 8k = 10 + 2/k + 8k$.Using the $AM$-$GM$ inequality: $\frac{2/k + 8k}{2} \geq \sqrt{(2/k)(8k)} = \sqrt{16} = 4$.So,$2/k + 8k \geq 8$.The minimum value of $OA+OB = 10 + 8 = 18$.

Let a variable line passing through the centre of the circle $x^2+y^2-16x-4y=0$ meet the positive coordinate axes at points $A$ and $B$. Then the minimum value of $OA+OB$,where $O$ is the origin,is equal to

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