Two equal circles intersect such that each passes through the centre of the other. If the length of the common chord of the circles is 10√3 cm, then what is the diameter of the circle?a)10 cmb)15 cmc)20 cmd)30 cmCorrect answer is option 'C'. Can you explain this answer?

Question

Two equal circles intersect such that each passes through the centre of the other. If the length of the common chord of the circles is 10&radic;3 cm, then what is the diameter of the circle?a)10 cmb)15 cmc)20 cmd)30 cmCorrect answer is option 'C'. Can you explain this answer?

Answer

Let the radius of each circle be $r$. Since each circle passes through the center of the other, we know that the distance between the centers is $2r$.

[asy]
size(100);
pair A,B,C,D,E,F;
A=(0,0);
B=(2,0);
C=(1,sqrt(3));
D=(1,0);
E=(1,1.732);
F=(1,-1.732);
draw(circle(A,2));
draw(circle(B,2));
draw((-2,0)--(4,0),Arrows);
draw((0,-2)--(0,4),Arrows);
draw(C--D--F--cycle);
draw((1,0)--E,dashed);
label('$r$',(0.5,0),S);
label('$r$',(2.5,0),S);
label('$2r$',(1,sqrt(3)/2),N);
[/asy]

Let $O_1$ and $O_2$ be the centers of the circles, and let $AB$ be the common chord of the circles. Let $M$ be the midpoint of $AB$, and let $OM$ be the altitude from $O$ to $AB$. Then, $OM=\frac12 AB=5$. Also, $O_1MO_2$ is a right triangle with hypotenuse $O_1O_2=2r$ and altitude $OM=5$. Therefore, by the Pythagorean theorem, we have \begin{align*}
r^2&=(O_1M)^2\
&=(O_1O_2)^2-(OM)^2\
&=4r^2-25.
\end{align*} Solving for $r$ gives $r=\sqrt{\frac{25}{3}}$. Therefore, the area of each circle is $\pi r^2=\boxed{\frac{25\pi}{3}}$.

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