Question

Let ‘C’ be a point on a circle with centre ‘O’ and radius ‘r’. Chord AB of length ‘r’ is parallel to OC. Let the line AO cut the circle in ‘E’ and the tangent to the circle at ‘C’ in ‘F’. If the chord BE cuts OC in L and AL cuts CF in M, then a possible value of ratio $\frac{CF}{CM}$ equals

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4



$OCF\sim ABE$
$\frac{OC}{AB}=\frac{CF}{BE}$
$ABL\sim LCM$
$\frac{AB}{LC}=\frac{BL}{CM}=\frac{\frac{BE}{2}}{CM}$
$\frac{r}{r-OL}=\frac{CF}{2CM}$
$\frac{CF}{2CM}=\frac{2r}{r-\frac{1}{2}AB}=\frac{2r}{r/2}=4$