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(B) The circle is $x^2+y^2-12x-16y=0$. Its center $C$ is $(6, 8)$ and radius $r = \sqrt{6^2+8^2} = 10$.Let $P$ be the point of intersection of the tan

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$5(2 \pm 5 \sqrt{2})$

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(B) The circle is $x^2+y^2-12x-16y=0$. Its center $C$ is $(6, 8)$ and radius $r = \sqrt{6^2+8^2} = 10$.Let $P$ be the point of intersection of the tangents. The angle between the tangents is $60^{\circ}$,so the angle between the line joining the center to $P$ and the tangent is $30^{\circ}$.In $	riangle PAC$,$\sin(30^{\circ}) = \frac{AC}{PC} = \frac{10}{PC} = \frac{1}{2}$,so $PC = 20$.The distance from the center $(6, 8)$ to the chord $5x-5y+k=0$ is $d = \frac{|5(6)-5(8)+k|}{\sqrt{5^2+(-5)^2}} = \frac{|k-10|}{5\sqrt{2}}$.In $	riangle PAC$,the distance from $C$ to the chord $AB$ is $d = r \cos(30^{\circ}) = 10 	imes \frac{\sqrt{3}}{2} = 5\sqrt{3}$.Wait,the angle between the tangents is $60^{\circ}$,so the angle $\angle APC = 30^{\circ}$. In $	riangle PAC$,$\angle PAC = 90^{\circ}$,so $\angle PCA = 60^{\circ}$.Thus,$d = r \cos(60^{\circ}) = 10 	imes \frac{1}{2} = 5$.Therefore,$\frac{|k-10|}{5\sqrt{2}} = 5 \Rightarrow |k-10| = 25\sqrt{2}$.$k-10 = \pm 25\sqrt{2} \Rightarrow k = 10 \pm 25\sqrt{2} = 5(2 \pm 5\sqrt{2})$.

Column $I$	Column $II$
$(A)$ Two intersecting circles	$(p)$ have a common tangent
$(B)$ Two mutually external circles	$(q)$ have a common normal
$(C)$ Two circles,one strictly inside the other	$(r)$ do not have a common tangent
$(D)$ Two branches of a hyperbola	$(s)$ do not have a common normal

If the angle between the tangents drawn to the circle $x^2+y^2-12x-16y=0$ at the points where the line $5y=5x+k$ cuts the circle is $60^{\circ}$,then the value of $k$ is

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