The circle x^2+y^2-4x-8y+16=0 rolls along the | vedclass

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(A) The given circle is $x^2+y^2-4x-8y+16=0$. Its center is $C(2, 4)$ and radius $r = \sqrt{2^2+4^2-16} = 2$.Let $P(2+\sqrt{3}, 3)$ be the point of ta

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$x^2+y^2-6x-2(4+\sqrt{3})y+(24+8\sqrt{3})=0$

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(A) The given circle is $x^2+y^2-4x-8y+16=0$. Its center is $C(2, 4)$ and radius $r = \sqrt{2^2+4^2-16} = 2$.Let $P(2+\sqrt{3}, 3)$ be the point of tangency.The slope of the radius $CP$ is $m_{CP} = \frac{3-4}{(2+\sqrt{3})-2} = \frac{-1}{\sqrt{3}}$.The tangent is perpendicular to the radius,so the slope of the tangent is $m_t = \sqrt{3}$.Since the circle rolls along the tangent by $2$ units,the new center $C'$ is at a distance of $2$ units from $C$ in the direction of the tangent.The angle of the tangent with the $x$-axis is $	heta = 	an^{-1}(\sqrt{3}) = 60^{\circ}$.The new center $C'$ is $(2+2\cos 60^{\circ}, 4+2\sin 60^{\circ}) = (2+2(1/2), 4+2(\sqrt{3}/2)) = (3, 4+\sqrt{3})$.The radius remains $r=2$. The equation of the new circle is $(x-3)^2 + (y-(4+\sqrt{3}))^2 = 2^2$.Expanding this: $x^2-6x+9 + y^2-2(4+\sqrt{3})y + (4+\sqrt{3})^2 = 4$.$x^2+y^2-6x-2(4+\sqrt{3})y + 9 + 16 + 3 + 8\sqrt{3} - 4 = 0$.$x^2+y^2-6x-2(4+\sqrt{3})y + (24+8\sqrt{3}) = 0$.

The circle $x^2+y^2-4x-8y+16=0$ rolls along the tangent drawn to it at $(2+\sqrt{3}, 3)$ by $2$ units. The equation of the circle in the new position is

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