Area of a circle passing through two vertices of a parallelogram touching one edge.

Question

For reference:

Let $ABCD$ be a parallelogram, $AB = 6, BC= 10$ and $AC = 14$ ; traces a circle passing through the vertices $C$ and $D$ with $AD$ being tangent and this circle is $BC$ a secant. Calculate the area of ​​the circle. (Answer: $12\pi$)

My progress:

$AD^2 =AI\cdot AC \implies 10^2=AI\cdot14$

$ \implies AI = \dfrac{50}{7} ,\ IC = \dfrac{48}{7}$

In $\triangle ABC$

$14^2=6^2+10^2-2\cdot6\cdot10\cdot\cos \angle B\implies \cos \angle B =-\dfrac{1}{2}\therefore \angle B =120^o $

....I can't see many options????

Answer 1

Drop the perp from $A$ to $BC$ and call the foot $E$.

Let $BE=x$, then using Pythagoras' theorem, $$6^2-x^2=14^2-(10+x)^2\implies x=3.$$

Therefore in right triangle $ABE$, $\angle BAE=30^\circ$ and so is $\angle CDO$.

Let $M$ be the midpoint of side $CD$, then considering $\triangle ODM$, $OD$, or, radius of the circle is $2\sqrt3$.

Hence, the area of circle is $12\pi$.

Answer 2

Comment: My understanding by "circle is a BC secant" is that the center of circle O lies on bisector of $\angle BCD$. You can easily find the radius equal to $34.64$ hence the area of circle is:

$s=3.464^2\pi\approx 12\pi$

Answer 3

The center $(x,y)$ of the circle has obvious $x=10$; the ordinate $y$ is given by the intersection of lines $x=10$ and the perpendicular to the segment $DC$ at its midpoint (and this ordinate is clearly the radius also by the tangency).

$► \cos(\angle{ABC})=-\dfrac12\Rightarrow \angle{BAD}=60^{\circ}$.

It follows:

$►$ Coordinates $D=(10,0),C=(13,3\sqrt3)$, midpoint of $DC=(11.5,1.5\sqrt3)$

$►$ Line $DC$ has pente $\sqrt3$ then the perpend. has equation $2y-3\sqrt3=-\dfrac{1}{\sqrt3}(2x-23)$ which gives for $x=10$ the value $y=2\sqrt3$.

$►$ Finally the area is $\pi r^2=12\pi$.