 Let us denote the radius of circle with center $A$ as $r_{1}$ and circle with center $C$ as $r_{2}$, and $AC=d$ as the distance between the centers. Now note by SAS criteria (prove $\triangle ADC$ and $\triangle ABC$ are congruent by SSS and ...) that $\triangle ABE$ and $\triangle ADE$ are congruent. Thus $\angle AEB = \angle AED$, but since $\angle AEB + \angle AED = 180^{\circ}$, we have $\angle AEB = \angle AED = 90^{\circ}$. Similarly, $\angle BEC = \angle DEC= 90^{\circ}$. Thus, we can see that $BE$ is the height of $\triangle ABC$, and $DE$ is the height of $\triangle ADC$, with base $AC$. So, $$BD=BE+DE=\frac{2(ABC)}{AC} + \frac{2(ADC)}{AC}=\frac{4}{d}(ABC)$$ Where $(ABC)$ and $(ADC)$ represents the area of the respective triangles, which we can calculate in terms of $r_{1},r_{2},d$ using the Herons formula. This is the general formula which gives the length of the common chord, $BD$. I leave the proof of the above formula above as a exercise for you. However, your formula was for a special case, when $\angle ABC = \angle ADC= 90^{\circ}$. This simplifies the area of the triangles and gives: $$BC=\frac{4}{d}(ABC)=\frac{2r_{1}r_{2}}{d}$$ Since, in your first question, the pair $(r_{1},r_{2},d)$ was $(15,20,25)$, which is a Pythagorean triple, so the condition $\angle ABC = \angle ADC= 90^{\circ}$ held, and so did your formula. Now lets come to the second question, with a new diagram: I leave it as a exercise to you to prove that $r_{1}=r_{2}=AB=AC=BC$. Hence, $ABC$ is an equilateral triangle with $\angle ABC = 60^{\circ} \neq 90^{\circ}$. Hence your formula does not hold, and we must return to the general formula: $$AD=\frac{4}{d}(ABC)=\frac{4}{d}\frac{\sqrt{3}(AC)^2}{4}=\frac{\sqrt{3}(r_{1})^2}{r_{1}}=\sqrt{3}r_{1}$$ What I essentially did in the second last step, was to use the Pythagoras theorem, like the way you must have did, to find the altitude in terms of the side, $r_{1}$, and use it to find the area of the triangle, or we can directly find $AD$ from the height. Let the circles' centers be A and B. Draw the common chord CD. Draw radii AC and BC. We know that So far so good. The line that connects the two centers will cut the chord into 2 equal lengths and be perpendicular to it. Let the length of the chord be 2L. Then the line connecting the centers cuts that chord into 2 segments, each L long. So Now you have 2 sets of right triangles. Both sets have one leg of length L. One set has a hypotenuse of 15; the other of 13. The sum of the two third legs is 14. Let the length of the thrid leg on the larger triangle be x. Then the lenght of the third leg on the shorter one is (14-x). Use Pythagorean theorem twice |
Two circles intersect and have a common chord. the radii of the circles are 13 and 15
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