Point A lies in the exterior of odot(P, 10). | vedclass

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(A) Given that point $A$ is in the exterior of the circle with center $P$ and radius $r = 10$. Since the line from $A$ touches the circle at $B$,$AB$

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) Given that point $A$ is in the exterior of the circle with center $P$ and radius $r = 10$. Since the line from $A$ touches the circle at $B$,$AB$ is a tangent to the circle. $A$ radius drawn to the point of tangency is perpendicular to the tangent line. Therefore,$\angle PBA = 90^{\circ}$. In the right-angled triangle $	riangle PBA$,by the Pythagorean theorem: $PA^2 = PB^2 + AB^2$. Given $PA = 26$ and $PB = r = 10$. $26^2 = 10^2 + AB^2$. $676 = 100 + AB^2$. $AB^2 = 676 - 100 = 576$. $AB = \sqrt{576} = 24$.

Point $A$ lies in the exterior of $\odot(P, 10)$. $A$ line from $A$ touches the circle at $B$. If $PA = 26$,then find the length of $AB$.

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