Suppose two tangents PA and PB are drawn to t | vedclass

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(A) The distance $PC$ between the center $C(1, 2)$ and the point $P(16, 7)$ is given by:$PC = \sqrt{(16-1)^2 + (7-2)^2} = \sqrt{15^2 + 5^2} = \sqrt{22

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$5$

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(A) The distance $PC$ between the center $C(1, 2)$ and the point $P(16, 7)$ is given by:$PC = \sqrt{(16-1)^2 + (7-2)^2} = \sqrt{15^2 + 5^2} = \sqrt{225 + 25} = \sqrt{250}$.The quadrilateral $PACB$ consists of two congruent right-angled triangles $	riangle PAC$ and $	riangle PBC$. The area of $PACB$ is $2 	imes 	ext{Area}(	riangle PAC) = 2 	imes (\frac{1}{2} 	imes AP 	imes r) = AP 	imes r = 75$.Let $AP = x$,so $x = \frac{75}{r}$.In right-angled triangle $	riangle PAC$,by the Pythagorean theorem:$x^2 + r^2 = PC^2$$(\frac{75}{r})^2 + r^2 = 250$$\frac{5625}{r^2} + r^2 = 250$$r^4 - 250r^2 + 5625 = 0$.Let $u = r^2$,then $u^2 - 250u + 5625 = 0$.$(u - 225)(u - 25) = 0$.Thus,$r^2 = 225$ or $r^2 = 25$.If $r^2 = 225$,then $r = 15$. Then $x = \frac{75}{15} = 5$. In this case,$PC^2 = 15^2 + 5^2 = 250$,which is consistent.If $r^2 = 25$,then $r = 5$. Then $x = \frac{75}{5} = 15$. In this case,$PC^2 = 5^2 + 15^2 = 250$,which is also consistent.Since $5$ is an option,the radius is $5$.

Suppose two tangents $PA$ and $PB$ are drawn to the circle centered at $C(1, 2)$ from the point $P(16, 7)$. If the area of the quadrilateral $PACB$ is $75$ square units,then the radius of the circle is:

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