Question

Find the equation of tangent to the circle $x^2+y^2-4x-8y+16=0$ at the point $(2+\sqrt{3},3)$. If the circle rolls up along this tangent by $2$ units then find its equation in the new position.

Answer 1

The given circle is $(2,4),2$
The equation of the tangent at the given point is found to be $\sqrt{3}x-y-2\sqrt{3}=0$, whose slope is $\sqrt{3}$ i.e. it makes an angle ${60}^{\circ }$ with x-axis. After the circle rolls up along the tangent through a distance $2$, its centre will move from $C_1$ to $C_2$ whose co-ordinates we have to find. Now $C_1{C}_2$ makes an angle of ${60}^{\circ }$ and passes through $C_1(2,4)$ and $C_2$ is at a distance of $2$ units from $C_1$
$\therefore \frac{x-2}{\cos {60}^{\circ }}=\frac{y-4}{\sin {60}^{\circ }}=r=2$ for $C_2$
$\therefore C_2=(2+2.\frac{1}{2},4+2.\frac{\sqrt{3}}{2})=(3,4+\sqrt{3})$
Hence the equation of circle in new position is $(x-3{)}^2+(y-4-\sqrt{3}{)}^2=2^2$.