1966 AHSME Problems/Problem 40

Problem

In this figure $AB$ is a diameter of a circle, centered at $O$ , with radius $a$ . A chord $AD$ is drawn and extended to meet the tangent to the circle at $B$ in point $C$ . Point $E$ is taken on $AC$ so the $AE=DC$ . Denoting the distances of $E$ from the tangent through $A$ and from the diameter $AB$ by $x$ and $y$ , respectively, we can deduce the relation:

$\text{(A) } y^2=\frac{x^3}{2a-x} \quad \text{(B) } y^2=\frac{x^3}{2a+x} \quad \text{(C) } y^4=\frac{x^2}{2a-x} \\ \text{(D) } x^2=\frac{y^2}{2a-x} \quad \text{(E) } x^2=\frac{y^2}{2a+x}$

Solution

$\fbox{A}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Problem 40
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1966 AHSME Problems/Problem 40

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