Point A lies in the exterior of odot(O, 8). A | vedclass

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(B) Given that the circle has a center $O$ and radius $r = 8$. Point $A$ is in the exterior of the circle. $A$ line through $A$ touches the circle at

Vedclass · Accepted Answer

$17$

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(B) Given that the circle has a center $O$ and radius $r = 8$. Point $A$ is in the exterior of the circle. $A$ line through $A$ touches the circle at $B$,which means $AB$ is a tangent to the circle at point $B$. By the property of tangents,the radius $OB$ is perpendicular to the tangent $AB$ at the point of contact $B$. Therefore,$	riangle OBA$ is a right-angled triangle with $\angle OBA = 90^{\circ}$. In $	riangle OBA$,by the Pythagorean theorem: $OA^2 = OB^2 + AB^2$ Given $OB = r = 8$ and $AB = 15$. $OA^2 = 8^2 + 15^2$ $OA^2 = 64 + 225$ $OA^2 = 289$ $OA = \sqrt{289} = 17$. Thus,the length of $OA$ is $17$.

Point $A$ lies in the exterior of $\odot(O, 8)$. $A$ line through $A$ touches the circle at $B$. If $AB = 15$,then find $OA$.

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