Question

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 that $AE=DC.$ If the coordinates of $E$ are $(x,y),$ then:

$y^2=\dfrac{x^3}{2a-x}$

$y^2=\dfrac{x^3}{2a+x}$

$y^4=\dfrac{x^2}{2a-x}$

$x^2=\dfrac{y^2}{2a-x}$

$x^2=\dfrac{y^2}{2a+x}$

Answer: A

Concepts:circle coordinate geometry similarity tangent line

Difficulty rating: 2060

Small Hint:

Drop perpendiculars from $E$ and $D$ to diameter $AB$

Big Hint:

Use $AE=DC$ between the two parallel tangents, then combine the altitude theorem with similar triangles

Solution:

Drop perpendiculars $EM$ and $DN$ to $AB.$ Since $AE=DC$ and the tangents through $A,B$ are parallel, their projections give $NB=x.$ Hence $AN=2a-x.$ In right triangle $ADB,$ the altitude theorem gives $DN^2=x(2a-x).$ Similar triangles $AME$ and $AND$ give $\frac{DN}{2a-x}=\frac{y}{x},$ so $DN=\frac{y(2a-x)}{x}.$ Substitution and cancellation yield $y^2=\frac{x^3}{2a-x}.$

Therefore, the correct answer is A.

1966 AMC 12 Problem 40

Answer: A

Difficulty rating: 2060

Small Hint:

Big Hint:

Solution:

Problem 40 in Other Years