Two circles with centres A and B touch each other internally. Another circle touches the larger circle internally at the point X and the smaller circle externally at the point Y. If O be the centre of that circle, prove that (OA + BO) is constant.

Explanation

Let the circle with centre A touches the circle with centre B internally at the circle with centre B internally at the point C. We know that if two circles touches each other either externally or internally, then the centres of the circles and the point of contact lie on the same straight line.
`therefore` the points A, O, X, A, B, C and B, Y, O lie on the same straight line each.
Now, `AO + BO + BY+OY` [as per. figure]
`=OA + OY +BY`
`=OA + OX + BY [because OY =OX =` radii of same circle]
`=AX + BY` [as per figure `AO +OX =AX]`
= Radius of hte larger circle + Radius of the smaller circle.
= Constant `[because` both the circles are fixed ]
Hence `(AO+BO)` = constant (Proved)