overline{AB} is a diameter of odot(O, 15). A | vedclass

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(D) In $\odot(O, 15)$,$OA = OB = 15$ (radius) and $AB = 30$ (diameter).In $\odot(O, 9)$,$OD = 9$ (radius).Since $BD$ is a tangent to $\odot(O, 9)$ at

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$18$

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(D) In $\odot(O, 15)$,$OA = OB = 15$ (radius) and $AB = 30$ (diameter).In $\odot(O, 9)$,$OD = 9$ (radius).Since $BD$ is a tangent to $\odot(O, 9)$ at $D$,we have $\overline{OD} \perp \overline{BD}$. Thus,$	riangle ODB$ is a right-angled triangle with $\angle ODB = 90^{\circ}$.Since $AB$ is a diameter of $\odot(O, 15)$,any angle inscribed in a semicircle is a right angle. Therefore,$\angle ACB = 90^{\circ}$.Now,consider $	riangle ODB$ and $	riangle ACB$:$1$. $\angle ODB = \angle ACB = 90^{\circ}$$2$. $\angle DBO = \angle CBA$ (Common angle)By $AA$ similarity criterion,$	riangle ODB \sim 	riangle ACB$.Therefore,the ratios of corresponding sides are equal:$\frac{AC}{OD} = \frac{AB}{OB}$$\frac{AC}{9} = \frac{30}{15}$$\frac{AC}{9} = 2$$AC = 9 	imes 2 = 18$.

$\overline{AB}$ is a diameter of $\odot(O, 15)$. $A$ tangent is drawn from $B$ to $\odot(O, 9)$ which touches $\odot(O, 9)$ at $D$. $\overrightarrow{BD}$ intersects $\odot(O, 15)$ at $C$. Find $AC$.

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