Cheenta
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

Circle Problem | AMC 10A, 2006 | Problem 23

Try this beautiful problem from Geometry: Circle

Circle Problem - AMC-10A, 2006- Problem 23


Circles with centers $A$ and $B$ have radii 3 and 8 , respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at $E,$ and $A E=5 .$ What is $C D ?$

,

 i

  • $13$
  • $\frac{44}{3} $
  • $\sqrt{221}$
  • $\sqrt{255}$
  • \(\frac{55}{3}\)

Key Concepts


Geometry

Circle

Tangents

Check the Answer


Answer: $ \frac{44}{3}$

AMC-10 (2006) Problem 23

Pre College Mathematics

Try with Hints


Circle Problem

Given that Circles with centers $A$ and $B$ have radii 3 and 8 and $A E=5 .$.we have to find out \(CD\).So join \(BC\) and \(AD\).then clearly \(\triangle BCE\) and \(\triangle ADE\) are Right-Triangle(as \(CD\) is the common tangent ).Now \(\triangle BCE\) and \(\triangle ADE\) are similar.Can you proof \(\triangle BCE\) and \(\triangle ADE\)?

Can you now finish the problem ..........

Circle Problem

$\angle A E D$ and $\angle B E C$ are vertical angles so they are congruent, as are angles $\angle A D E$ and $\angle B C E$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $\triangle A C E \sim \triangle B D E$.

By the Pythagorean Theorem, line segment \(DE=4\)

Therefore from the similarity we can say that \(\frac{D E}{A D}=\frac{C E}{B C} \Rightarrow \frac{4}{3}=\frac{C E}{8}\) .

Therefore \(C E=\frac{32}{3}\)

can you finish the problem........

Therefore \(CD=CE+DE=4+\frac{32}{3}=\frac{44}{3}\)

Subscribe to Cheenta at Youtube


Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com