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If a tangent segment and a secant segment intersect outside a circle, then the square of the measure of the tangent segment equals the product of the measures of the secant segment and its external portion. x^2 = (x-2)*(2x-3) x = 1 or 6, I think
How would you do 18?
I have to go to bed. I'll check your answer in the morning. Good night!
((x-1)(x+5) = y^2) and ( (2y)^2 = (2x)(2x+6)) x = 5 and y = 2√10, I think. This: If a tangent segment and a secant segment intersect outside a circle, then the square of the measure of the tangent segment equals the product of the measures of the secant segment and its external portion. has to be used twice and the simultaneous equations solved.
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