Calculate the distance from O to AC in the figure below.

Question

If $T$ is a point of tangency, $(OB)(OC)=24$, $TM=MC$ and $AC=8$. Calculate the distance from O to AC.(Answer:$3$)

$\triangle OFT\sim \triangle CFO \implies \frac{CO}{r}=\frac{FO}{FT}=\frac{CF}{FO}\\ MT^2 =CM^2= DM(DM+2r)\\ \triangle MTO: MO^2 = MT^2++r^2 = CM^2+r^2\\ OB.OC = 24 \implies r.(r+DC)=24 $

Answer 1

Draw $OT$, $OA$, $BT$ and $OF\perp AC$

$OT\perp TM$ (rad $\perp $ tan)

First Solution :

Let $\angle MCT = \angle MTC = \alpha$ ($\angle$'s are equal opposite $=$ sides $MT=MC$)

then

$\angle TMO=2\alpha$, $\angle TOM = 90^{\circ} - 2 \alpha$

$\angle TBD= 45^{\circ} - \alpha$

and $\angle OTB = \angle TBD= 45^{\circ} - \alpha$ $(OB=OT = r)$

$\angle ATB= 45^{\circ}$ (straigth $\angle$)

$\angle AOB= 90^{\circ}$ ($\angle$ at centre $= 2\times \angle$ at circumference)

then

$OB\times OC=OA\times OC=AC\times OF$

$24=8\times OF$

$OF=3$

Secod Solution :

Let $\angle MCT = \angle MTC = \alpha$ ($\angle$'s are equal opposite $=$ sides $MT=MC$)

$\angle OTA = 90^{\circ} - \alpha$ (straigth $\angle$)

$\angle OAT = \angle OTA = 90^{\circ} - \alpha$ ($OA=OT=r$)

then $\angle AOC = 90^{\circ} $