Two circles from a distance

Geometry Level 2

B C BC is tangent to both a circle with center at A A and a circle with center at D D . The area of the circle with center at A A is 225 π 225\pi and the area of the circle with center at D D is 36 π 36\pi .

If B C = 16 BC=16 , find the distance between the centers of the two circles.

697 \sqrt{697} 1 7 π + 697 \dfrac{1}{7}\pi+\sqrt{697} 5 697 7 + 2 679 7 \dfrac{5\sqrt{697}}{7}+\dfrac{2\sqrt{679}}{7} 17425 49 + 27888 49 \sqrt{\dfrac{17425}{49}}+\sqrt{\dfrac{27888}{49}}

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1 solution

Consider the diagram. △ A B E ∼ △ D C E \triangle ABE \sim \triangle DCE , thus

B E A B = C E C D \dfrac{BE}{AB}=\dfrac{CE}{CD}

B E 15 = 16 − B E 6 \dfrac{BE}{15}=\dfrac{16-BE}{6}

6 B E = 240 − 15 B E 6BE=240-15BE

21 B E = 240 21BE=240

B E = 80 7 BE=\dfrac{80}{7}

It follows that

C E = 16 − 80 7 = 32 7 CE=16-\dfrac{80}{7}=\dfrac{32}{7}

Hence,

A D = A B 2 + B E 2 + C D 2 + C E 2 AD=\sqrt{AB^2+BE^2}+\sqrt{CD^2+CE^2}

A D = 1 5 2 + ( 80 7 ) 2 + 6 2 + ( 32 7 ) 2 AD=\sqrt{15^2+\left(\dfrac{80}{7}\right)^2}+\sqrt{6^2+\left(\dfrac{32}{7}\right)^2}

A D = 17425 49 + 2788 49 AD=\sqrt{\dfrac{17425}{49}}+\sqrt{\dfrac{2788}{49}}

A D = 5 697 7 + 2 697 7 AD=\dfrac{5\sqrt{697}}{7}+\dfrac{2\sqrt{697}}{7}

A D = 697 \boxed{AD=\sqrt{697}}

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