Finding the Length of the Common Chord of Intersecting Circles

This problem involves two equal circles that intersect in a specific way: each circle passes through the center of the other circle. We are given the radius of these circles and need to find the length of the common chord.

Understanding the Geometry of Intersecting Circles

Let the two equal circles be \(C_1\) and \(C_2\), and let their centers be \(O_1\) and \(O_2\), respectively. The radius of both circles is given as \(r = 18\) cm.

Since each circle passes through the center of the other, the distance between the centers \(O_1\) and \(O_2\) is equal to the radius, \(r\). So, \(O_1O_2 = 18\) cm.

The common chord is the line segment connecting the two points where the circles intersect. Let these intersection points be \(A\) and \(B\). The line segment \(AB\) is the common chord.

A key property of the common chord of two intersecting circles is that it is perpendicular to the line joining their centers, and it is bisected by this line. Let \(M\) be the point where the common chord \(AB\) intersects the line segment \(O_1O_2\).

Therefore, \(AB \perp O_1O_2\), and \(AM = MB\). Also, \(M\) is the midpoint of \(O_1O_2\) because the circles are equal and each passes through the other's center. This means \(O_1M = MO_2 = \frac{O_1O_2}{2} = \frac{r}{2}\).

Calculating Half the Length of the Common Chord using Pythagoras Theorem

Consider the triangle \(O_1MA\). This is a right-angled triangle with the right angle at \(M\).

• \(O_1A\) is the radius of circle \(C_1\), so \(O_1A = r = 18\) cm.
• \(O_1M\) is half the distance between the centers, so \(O_1M = \frac{r}{2} = \frac{18}{2} = 9\) cm.
• \(AM\) is half the length of the common chord \(AB\).

According to the Pythagorean theorem in triangle \(O_1MA\):

\( (O_1A)^2 = (O_1M)^2 + (AM)^2 \)

Substitute the known values:

\( (18)^2 = (9)^2 + (AM)^2 \)

\( 324 = 81 + (AM)^2 \)

Now, solve for \( (AM)^2 \):

\( (AM)^2 = 324 - 81 \)

\( (AM)^2 = 243 \)

Take the square root to find \(AM\):

\( AM = \sqrt{243} \)

Simplifying the Result and Finding the Common Chord Length

We need to simplify \(\sqrt{243}\). We can factorize 243:

\( 243 = 3 \times 81 \)

Since \(81 = 9^2\), we have:

\( \sqrt{243} = \sqrt{81 \times 3} = \sqrt{81} \times \sqrt{3} = 9\sqrt{3} \)

So, \(AM = 9\sqrt{3}\) cm.

The length of the common chord \(AB\) is twice the length of \(AM\):

\( AB = 2 \times AM = 2 \times 9\sqrt{3} = 18\sqrt{3} \) cm.

Comparing with the Given Options

Let's check the given options to see which one matches \(18\sqrt{3}\) cm.

• Option 1: \(6\sqrt{27}\) cm. Let's simplify \(\sqrt{27}\): \( \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3} \). So, \(6\sqrt{27} = 6 \times (3\sqrt{3}) = 18\sqrt{3}\) cm. This matches our calculated length.
• Option 2: \(\sqrt{3}\). This is not \(18\sqrt{3}\).
• Option 3: \(9\sqrt{27}\) cm. \(9\sqrt{27} = 9 \times (3\sqrt{3}) = 27\sqrt{3}\) cm. This is not \(18\sqrt{3}\).
• Option 4: \(3\sqrt{3}\) cm. This is not \(18\sqrt{3}\).

The length of the common chord is \(18\sqrt{3}\) cm, which is equivalent to \(6\sqrt{27}\) cm.

Revision Table: Common Chord Calculations

Additional Information: Geometry of Intersecting Circles

When two circles intersect, the line connecting their centers is the perpendicular bisector of their common chord. This is a fundamental property used in geometry problems involving intersecting circles.

In the special case where two equal circles each pass through the center of the other:

• The distance between the centers is equal to the radius of the circles.
• The line segment connecting the centers and the two radii to an intersection point form an equilateral triangle. For example, triangle \(O_1AO_2\) would have sides \(O_1A = r\), \(O_2A = r\), and \(O_1O_2 = r\).
• The common chord passes through the midpoints of the radii connecting the centers to the intersection points (this is incorrect, the common chord bisects the line segment between the centers). The common chord is the altitude of the equilateral triangle formed by the centers and one intersection point, when considering the base as the line segment between the centers.
• The length of the altitude of an equilateral triangle with side 'a' is \(\frac{a\sqrt{3}}{2}\). In our case, the 'equilateral triangle' is \(O_1AO_2\) with side 'r'. The altitude from \(A\) to \(O_1O_2\) is \(AM\). So, \(AM = \frac{r\sqrt{3}}{2}\). The common chord \(AB = 2 \times AM = 2 \times \frac{r\sqrt{3}}{2} = r\sqrt{3}\).
• Using \(r=18\) cm, the common chord length is \(18\sqrt{3}\) cm. This confirms the result obtained using the Pythagorean theorem on the right triangle \(O_1MA\). The Pythagorean method (\(AM = \sqrt{r^2 - (r/2)^2} = \sqrt{\frac{3r^2}{4}} = \frac{r\sqrt{3}}{2}\)) is essentially deriving the altitude formula for this specific triangle configuration.

Understanding these geometric properties helps solve various problems related to intersecting circles and common chords.