Question

Drawtwo congruent circles with radii $\mathbf{6}$ each passing through the centre of other and to find length of their common chord.

Short answer

$\mathbf{6}\sqrt{\mathbf{3}}$

Step-by-step solution

Step 1. Given information.  

Radii of both the circles are $6$ and they pass through the centre of each other.

Step 2. Concept used.

Perpendicular from centre of circle to the chord bisects he chord.

Step 3. Solution. 

Two circles can be drawn as:

To find the length of the common chord $AB in \Delta \text{ADC}$:

$\begin{array}{c}AD=\sqrt{A{C}^2+C{D}^2}\\ AD=\sqrt{A{C}^2+{\left(\frac{DE}{2}\right)}^2}\\ 6=\sqrt{A{C}^2+{\left(\frac{6}{2}\right)}^2}\\ 6^2=A{C}^2+{\left(\frac{6}{2}\right)}^2\\ 36=A{C}^2+9\\ A{C}^2=27\\ AC=\sqrt{27}\\ AC=3\sqrt{3}\end{array}$

Perpendicular to any chord bisects the chord therefore,

$\begin{array}{l}AB=2AC\\ AB=2\times \left(3\sqrt{3}\right)\\ AB=6\sqrt{3}\end{array}$

Therefore, the answer is $\mathbf{6}\sqrt{\mathbf{3}}$.