Mensuration question about a hoop resting on a staircase

Question

I came across a question recently:

A hoop, as shown in the diagram, rests vertically at stair case. Note: AB = 12 cm, and BC = 8 cm. Find the radius of the hoop.
Figure (hand-made):

This is a question that I have no clue about. I thought that AB might be the radius but well from the diagram it seems no.

A thorough Google search revealed only one site which requires a login, which I don't want to do.

Can anyone please provide a explanatory answer.

Thanks a lot.

Answer 1

Let $Q$ be the center of the circle. Draw $QA$, and draw a horizontal line from $C$. Where this line meets $QA$, call it $J$. Then you should be able to get the answer by working with the triangle, $QJC$.

Answer 2

Let the radius be $r$, the center of the circle $O$, and draw the horizontal radius to the right. Drop (raise?) the perpendicular from $C$ to that radius, and call the intersection point $D$. Now $OCD$ is a right-triangle with hypotenuse $OC = r$, and legs $CD = r-8$ and $DO = 12$ (the center has to be directly above $A$).

Answer 3

The line through $A$ and $C$ has slope $8/12=2/3$, so the perpendicular bisector of $AC$ has slope $-3/2$. It will pass through the center of the circle directly above $A$. Taking $A$ as the origin, $(0,0)$, the midpoint of $AC$ is $(6,4)$, so the equation of the perpendicular bisector is $(y-4)=(-3/2)(x-6)$, so the center of the circle is at $(0,13)$, and hence the radius of the hoop is $13$.

Answer 4

Hint: find three similar triangles.

Answer 5

An analytic solution:

The equation of a circle, radius $R$ with centre at $(h, k)$ is:$$(x-h)^2+(y-k)^2=R^2$$If we take the point A as the origin, then the centre is at $(0, R)$ and the equation reduces to:$$(x)^2+(y-R)^2=R^2$$Note that the point $(12,8)$ lies on the circle and must satisfy the equation.