Two circles each of radius 36 cm are intersecting each other such that each circle is passing through the centre of the other circle. What is the length of common chord to the two circles ?
36 \(\sqrt{3}\) cm
Finding the Length of the Common Chord
This problem involves two identical circles that intersect in a specific way: each circle passes through the center of the other. This creates a symmetrical arrangement with some key geometric properties.
Understanding the Geometry of Intersecting Circles
Let's consider the two circles. Let the center of the first circle be \(C_1\) and the center of the second circle be \(C_2\). Both circles have a radius of 36 cm. The problem states that the first circle passes through \(C_2\) and the second circle passes through \(C_1\). This means the distance between the centers, \(C_1C_2\), is equal to the radius of both circles, which is 36 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 common chord is the line segment \(AB\).
Key Geometric Properties
- The line connecting the centers of the two circles (\(C_1C_2\)) is perpendicular to the common chord (\(AB\)).
- The line connecting the centers bisects the common chord. Let \(M\) be the point where \(C_1C_2\) intersects \(AB\). Then \(AM = MB\).
- Consider the triangle formed by the centers and one intersection point, for example, triangle \(C_1AC_2\). The sides of this triangle are \(C_1A\), \(C_2A\), and \(C_1C_2\).
Analyzing Triangle \(C_1AC_2\)
- \(C_1A\) is the radius of the first circle, which is 36 cm.
- \(C_2A\) is the radius of the second circle, which is also 36 cm.
- \(C_1C_2\) is the distance between the centers. Since each circle passes through the center of the other, \(C_1C_2\) is equal to the radius, which is 36 cm.
Therefore, triangle \(C_1AC_2\) is an equilateral triangle with side length 36 cm.
Similarly, triangle \(C_1BC_2\) is also an equilateral triangle with side length 36 cm.
Using the Properties to Find the Common Chord Length
As mentioned, the line \(C_1C_2\) is perpendicular to \(AB\) at point \(M\), and \(M\) is the midpoint of \(C_1C_2\). In triangle \(C_1AC_2\), \(AM\) is the altitude from \(A\) to the base \(C_1C_2\).
In an equilateral triangle, the altitude from a vertex to the opposite side is also the median. This means \(M\) is the midpoint of \(C_1C_2\).
The length of \(C_1C_2\) is 36 cm. So, \(C_1M = MC_2 = \frac{1}{2} \times C_1C_2 = \frac{1}{2} \times 36 = 18\) cm.
Applying the Pythagorean Theorem
Now consider the right-angled triangle \(C_1MA\). We know:
- Hypotenuse \(C_1A = 36\) cm (radius)
- Side \(C_1M = 18\) cm (half the distance between centers)
- Side \(AM\) is half the length of the common chord.
Using the Pythagorean theorem, \(C_1A^2 = C_1M^2 + AM^2\).
\(36^2 = 18^2 + AM^2\)
\(1296 = 324 + AM^2\)
\(AM^2 = 1296 - 324\)
\(AM^2 = 972\)
To find \(AM\), we take the square root of 972:
\(AM = \sqrt{972}\)
We can simplify the square root:
\(972 = 324 \times 3 = 18^2 \times 3\)
\(AM = \sqrt{18^2 \times 3} = 18\sqrt{3}\) cm.
Calculating the Common Chord Length AB
The common chord \(AB\) is twice the length of \(AM\).
\(AB = 2 \times AM\)
\(AB = 2 \times 18\sqrt{3}\)
\(AB = 36\sqrt{3}\) cm.
Thus, the length of the common chord is \(36\sqrt{3}\) cm.
Revision Table: Intersecting Circles and Common Chord
| Concept | Description | Relevance to Problem |
|---|---|---|
| Radius | Distance from center to any point on the circle. | Given as 36 cm. Defines circle size and distance \(C_1A\), \(C_2A\). |
| Distance between Centers | Distance between \(C_1\) and \(C_2\). | Equal to the radius (36 cm) because each circle passes through the other's center. This forms equilateral triangles. |
| Common Chord | Line segment connecting intersection points (\(A\) and \(B\)). | The length we need to find (\(AB\)). |
| Line of Centers | The line passing through \(C_1\) and \(C_2\). | Perpendicular bisector of the common chord. |
| Equilateral Triangle | Triangle with all sides equal. | \(C_1AC_2\) and \(C_1BC_2\) are equilateral triangles in this specific intersection case. |
| Pythagorean Theorem | \(a^2 + b^2 = c^2\) in a right triangle. | Used to find half the common chord length (\(AM\)) in right triangle \(C_1MA\). |
Additional Information: Common Chord Properties
When two circles intersect, the line segment connecting their intersection points is called the common chord. The line passing through the centers of the two circles has several important properties related to the common chord:
- The line of centers is always perpendicular to the common chord.
- The line of centers bisects the common chord. This means the common chord is divided into two equal halves by the line of centers.
- The length of the common chord depends on the radii of the two circles and the distance between their centers. For circles of unequal radii or different distances between centers, the calculation would vary.
In the special case where two circles of equal radius intersect, and each passes through the center of the other, the distance between centers is equal to the radius, and the triangles formed by the centers and intersection points are equilateral, simplifying the calculation significantly.
Similar Questions
If D is the midpoint of BC in △ABC and ∠A = 90⁰, then AD = _______.
AB is the diameter of a circle with centre O. P be a point on it. If ∠AOP = 95°, then ∠OBP __________.
In a circle, a 14 cm long chord is at 24 cm from the centre of the circle. Find the length of the radius of the circle.
If the area of a circle is 616 cm2 and a chord XY = 10 cm, then find the perpendicular distance from the center of the circle to the chord XY.
An arc of length 23.1 cm subtends an 18° angle at the centre. What is the area of the circle? [Use \(π = \frac{22}{7}\)]
Find the area of the sector of a circle with radius 5 cm and angle 60° (rounded off to one decimal).
A sector of a circle of radius 10 cm is formed at 60° angle at the centre. What will be its area (take π = 3.14)?
Two circles of radii 10 cm and 5 cm touch each other externally at a point A. PQ Is the direct common tangent of those two circles of centres O1 and O2 , respectively. The length of PQ is equal to:
Two circles of same radius 6 cm, intersect each other at P and Q. If PQ = 10 cm, then what is the distance between the centres of the two circles?
The distance between the centres of two circles of radii 2 cm and 6 cm is 5 cm. Find the length of the direct common tangent.
Important Questions from Circles, Chords and Tangents
If a tangent to a circle from a point P meets the circle at A with AP = 15 cm. Given that the radius of the circle is 8 cm, find the distance of P from the centre of the circle.
In a circle with a radius of 10 cm. XY and PQ are two parallel chords 12 cm and 16 cm in length, respectively. The two chords are situated on the opposite sides of the centre. The distance between the chords is:
Find the equation of the tangents to the circle x2 + y2 = 9 at x = 2.
If a chord of length 24 cm is at a distance of 5 cm from centre, then find the radius of the circle.
Let \( C_1 \) and \( C_2 \) be two circles which do not externally touch and intersect each other and \( O_1 \), and \( O_2 \) be the centers of the circles, respectively. Let AB be the common transverse tangent to the circles such that P, Q are the points of tangency respectively to \( C_1 \), \( C_2 \). Let R be the point of intersection of \( O_1 O_2 \) and AB. If \( \angle PO_1R = 60^\circ \), find \( \angle QO_2R \) and \( \angle QRO_2 \) respectively.
