Question

Two circles of radii 6 cm and 6 cm intersect each other. The distance between their centers is 8 cm. What is the length of their common chord?

• A. \(4\sqrt{5}\) cm
• B. \(6\sqrt{2}\) cm
• C. 8 cm
• D. 6 cm

Hint:

If a geometry problem's given numbers don't lead to any of the multiple-choice answers, check for a simple typo that might make the problem symmetric or easier. Assuming equal radii is a common correction.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have two circles intersecting at two points. Let the centers be C\(_1\) and C\(_2\), and the radii be r\(_1\) = 6 cm and r\(_2\) = 6 cm.
The distance between the centers is d = 8 cm. The common chord is the line segment connecting the two intersection points. The line connecting the centers (C\(_1\)C\(_2\)) is the perpendicular bisector of the common chord.
Step 2: Key Formula or Approach:
Let the common chord be AB, and let it intersect the line C\(_1\)C\(_2\) at point M. This forms two right-angled triangles, \(\triangle\)C\(_1\)MA and \(\triangle\)C\(_2\)MA. We can use the Pythagorean theorem on these triangles.
Let AM = h (half the length of the chord) and C\(_1\)M = x. Then C\(_2\)M = d - x = 8 - x.
In \(\triangle\)C\(_1\)MA: \(x^2 + h^2 = r_1^2\)
In \(\triangle\)C\(_2\)MA: \((8-x)^2 + h^2 = r_2^2\)
Step 3: Detailed Explanation:
The second radius was also meant to be 6 cm, i.e., r\(_1\) = 6 cm and r\(_2\) = 6 cm.
When the radii are equal, the problem becomes symmetric. The line connecting the centers bisects the distance, so \(x = d/2 = 8/2 = 4\) cm.
Now we use the Pythagorean theorem for one triangle:
\[ h^2 + x^2 = r_1^2 \] \[ h^2 + 4^2 = 6^2 \] \[ h^2 + 16 = 36 \] \[ h^2 = 36 - 16 = 20 \] \[ h = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \] The length of the common chord is \(2h\).
\[ \text{Length} = 2 \times (2\sqrt{5}) = 4\sqrt{5} \text{ cm} \] This result matches option (A).
Step 4: Final Answer
The length of the common chord is \(4\sqrt{5}\) cm.