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

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????

• Which circular region? Or the area of the whole circle? Jan 18 at 15:32
• @peterwhy whole circle........... Jan 18 at 15:36
• (without much optimisation) Let the centre of the circle be $O$. With $\cos \angle B$ known, find $\cos \angle ODC$, then find the radius $OD$ using that angle and the chord $CD$. Jan 18 at 15:43
• @petaarantes, what do you mean by this circle is BC a secant? Jan 18 at 15:59
• $\angle ODC=\angle ADC-\angle ADO = \angle B-90$, and triangle $ODC$ is isosceles with base length $6$. Jan 18 at 16:20

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 $$12\pi$$.

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 $$\pi r^2=12\pi$$.

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:

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