Find the value of $x$ where a tangent and secant meet outside the circle.

\sqrt{75}

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\sqrt{35}

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\sqrt{95}

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\sqrt{55}

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Solution

The correct option is C $\sqrt{75}$
The product of segments of secant and its external segment is equal to the square of length of the tangent .
$(A Y)^{2} = (B C + C Y) \times C Y$
$(A Y)^{2} = (10 + 5) \times 5$
$(A Y)^{2} = 75$
$A Y = \sqrt{75}$

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