Finding the Distance Between Centers of Intersecting Circles

The problem asks us to find the distance between the centers of two identical circles that intersect each other. We are given the radius of the circles and the length of their common chord.

Understanding the Geometry

When two circles intersect at two points, say P and Q, the line segment PQ is called the common chord. A key property of intersecting circles is that the line segment joining their centers is the perpendicular bisector of the common chord. This property is crucial for solving this problem.

Let the centers of the two circles be $O_1$ and $O_2$. The radius of each circle is given as 6 cm. So, $O_1P = O_1Q = O_2P = O_2Q = 6$ cm. The common chord PQ has a length of 10 cm.

Let M be the point where the line segment $O_1O_2$ intersects the common chord PQ. Since $O_1O_2$ is the perpendicular bisector of PQ, M is the midpoint of PQ, and the line $O_1O_2$ is perpendicular to PQ. Therefore, $\angle O_1MP = 90^\circ$ and $\angle O_2MP = 90^\circ$.

Because M is the midpoint of PQ, we have:

$\text{PM} = \text{MQ} = \frac{\text{PQ}}{2} = \frac{10}{2} = 5$ cm.

Using the Pythagorean Theorem

Consider the triangle $\triangle O_1MP$. This is a right-angled triangle with the right angle at M. We know the length of the hypotenuse $O_1P$ (which is the radius, 6 cm) and the length of one leg PM (which is 5 cm). We need to find the length of the other leg $O_1M$.

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. For $\triangle O_1MP$:

$(O_1P)^2 = (O_1M)^2 + (PM)^2$

Substitute the known values:

$(6)^2 = (O_1M)^2 + (5)^2$

$36 = (O_1M)^2 + 25$

Now, solve for $(O_1M)^2$:

$(O_1M)^2 = 36 - 25$

$(O_1M)^2 = 11$

Taking the square root of both sides to find $O_1M$:

$O_1M = \sqrt{11}$ cm.

Similarly, consider the triangle $\triangle O_2MP$. This is also a right-angled triangle with the right angle at M. The hypotenuse is $O_2P$ (radius, 6 cm), and one leg is PM (5 cm). Using the Pythagorean theorem for $\triangle O_2MP$:

$(O_2P)^2 = (O_2M)^2 + (PM)^2$

$(6)^2 = (O_2M)^2 + (5)^2$

$36 = (O_2M)^2 + 25$

$(O_2M)^2 = 11$

$O_2M = \sqrt{11}$ cm.

Since the circles have the same radius, the distance from each center to the midpoint of the common chord is the same, i.e., $O_1M = O_2M$.

Calculating the Distance Between Centers

The distance between the centers $O_1$ and $O_2$ is the length of the segment $O_1O_2$. This segment is formed by $O_1M$ and $MO_2$ (which is the same as $O_2M$) along a straight line.

Distance $O_1O_2 = O_1M + O_2M$

Distance $O_1O_2 = \sqrt{11} + \sqrt{11}$

Distance $O_1O_2 = 2\sqrt{11}$ cm.

Therefore, the distance between the centers of the two circles is $2\sqrt{11}$ cm.

Summary of Calculation Steps

• Identify the given information: Radius (r) = 6 cm, Common chord (PQ) = 10 cm.
• Recognize that the line connecting centers bisects the common chord perpendicularly.
• Calculate half the length of the common chord: PM = 10/2 = 5 cm.
• Use the Pythagorean theorem in the right triangle formed by a radius, half the common chord, and half the distance between centers. Let this half distance be x. $r^2 = x^2 + (\text{PM})^2$.
• Substitute values: $6^2 = x^2 + 5^2$.
• Solve for $x$: $36 = x^2 + 25 \Rightarrow x^2 = 11 \Rightarrow x = \sqrt{11}$ cm.
• The total distance between centers is $2x$ (since both circles have the same radius). Distance = $2 \times \sqrt{11} = 2\sqrt{11}$ cm.

The calculated distance matches one of the provided options.

Revision Table: Intersecting Circles Properties

Additional Information: Geometry of Intersecting Circles

The geometry of intersecting circles provides interesting problems and properties. Understanding these properties is essential for solving problems related to their common chord and the distance between their centers.

• If two circles intersect at two distinct points, the distance between their centers must be less than the sum of their radii and greater than the absolute difference of their radii. In our case, radii are 6 and 6. Sum = 12, Difference = 0. Distance $2\sqrt{11}$. $\sqrt{11}$ is between 3 and 4 (since $3^2=9$ and $4^2=16$), so $2\sqrt{11}$ is between 6 and 8. This is indeed less than 12 and greater than 0, confirming that intersection is possible.
• When the circles have the same radius, the line of centers bisects the common chord and also the angle formed by the radii to the intersection points from each center (forming congruent triangles like $\triangle O_1PQ$ and $\triangle O_2PQ$).
• The area of the region common to both circles can be calculated using the radii and the distance between centers, although this is a more advanced topic.

These principles are fundamental in geometry and are often applied in coordinate geometry and other areas of mathematics and physics.