Two circles of radii 7 cm and 5 cm intersect each other at A and B, and the distance between their centres is 10 cm. The length (in cm) of the common chord AB is:
Answer & explanation
Correct answer: option 2
.PNG)
Let us consider that , O'C = a cm & CO = 10 - a
In a right angled triangle AO'C
O'A² = O'C² + AC²
7² = a² + AC²
AC² = 49 - a²
In a right angled triangle AOC
OA² = OC² + AC²
5² = (10-a)² + AC²
AC² = 25 - (10-a)²
Now,
49 - a² = 25 - (10-a)²
49 - a² = 25 - 100 - a² + 20a
a = \(\frac{31}{5}\)
Now, AC² = 49 - a²
= 49 - (\(\frac{31}{5}\))²
= 49 - \(\frac{ 961}{25}\)
= \(\frac{ 264}{25}\)
AC = $\frac{2 \sqrt{66}}{5}$
We know AB = 2AC
= 2 x $\frac{2 \sqrt{66}}{5}$
= $\frac{4 \sqrt{66}}{5}$ cm