Question

Chords $\overline{AB}$ and $\overline{CD}$ of circle O intersect at point E. If $CE=3,ED=12$, and $AE$ is $5$ units longer than $EB,AB=$

• A. $4$
• B. $9$
• C. $11$
• D. $13$
• E. $18$

Answer 1

Answer: (D)

$13$

When chords intersects inside a circle the product of segment formed by the circle are equal.

$\therefore \left(AE\right)\left(EB\right)=\left(CE\right)\left(ED\right)$

Let the length of $EB$ is $x$

So according to the question $AE=x+5$

$\Rightarrow (x+5)\left(x\right)=\left(3\right)\left(12\right)$

$\Rightarrow x^2+5x=36$

$\Rightarrow x^2+5x-36=0$

$\Rightarrow x^2+9x-4x-36=0$

$\Rightarrow x(x+9)-4(x+9)=0$

$\Rightarrow (x+9)(x-4)=0$

$\Rightarrow x=-9,4$

Chord length is not negative, so $EB=4$

$\therefore AE=4+5=9$

$AB=AE+EB=9+4=13$