ABCDA is a con-cyclic quadrilateral of a circle ABCD with radius r and centre at O. If AB is the diameter and CD is parallel and half of AB and if the circle completes one rotation about the centre O, then the locus of the middle point of CD is a circle of radius:
√3r/2
Finding the Locus Radius of the Midpoint of CD
Let's analyze the given problem step by step to find the locus of the midpoint of the chord CD.
We are given a concyclic quadrilateral ABCD in a circle with center O and radius r. AB is the diameter of the circle, which means its length is \(2r\) and it passes through the center O. Points A and B are on the circle.
CD is a chord parallel to the diameter AB, and its length is half the length of AB. So, \(CD = \frac{1}{2} AB = \frac{1}{2} (2r) = r\). Points C and D are on the circle.
We need to find the locus of the middle point of CD when the circle (and thus the quadrilateral) completes one rotation about the center O.
Understanding the Geometry Setup
Let's place the center of the circle O at the origin (0, 0) in a coordinate system. Since AB is the diameter and parallel to CD, we can align AB with the x-axis. Thus, the coordinates of A and B are (-r, 0) and (r, 0) respectively.
Since CD is parallel to AB, CD is a horizontal chord. Let the equation of the circle be \(x^2 + y^2 = r^2\). Let the y-coordinate of the points C and D be \(y_0\). Since C and D are on the circle, their coordinates are \((x_C, y_0)\) and \((x_D, y_0)\) such that \(x_C^2 + y_0^2 = r^2\) and \(x_D^2 + y_0^2 = r^2\). This implies \(x_C^2 = x_D^2\), so \(x_D = -x_C\) (since C and D are distinct points).
Locating the Chord CD
The length of the chord CD is \(|x_D - x_C|\). We know \(CD = r\). So, \(|x_D - x_C| = |-x_C - x_C| = |-2x_C| = 2|x_C|\).
Given \(CD = r\), we have \(2|x_C| = r\), which means \(|x_C| = r/2\). We can choose \(x_C = -r/2\) and \(x_D = r/2\).
Now substitute \(x_C = -r/2\) into the circle equation \(x_C^2 + y_0^2 = r^2\):
\((-r/2)^2 + y_0^2 = r^2\)
\(r^2/4 + y_0^2 = r^2\)
\(y_0^2 = r^2 - r^2/4 = \frac{4r^2 - r^2}{4} = \frac{3r^2}{4}\)
\(|y_0| = \sqrt{\frac{3r^2}{4}} = \frac{\sqrt{3}r}{2}\)
So, the horizontal chord CD, which is parallel to AB and has length \(r\), must be located at a y-coordinate of either \(\frac{\sqrt{3}r}{2}\) or \(-\frac{\sqrt{3}r}{2}\). Let's consider the case where \(y_0 = \frac{\sqrt{3}r}{2}\). The coordinates of C and D can be \((-\frac{r}{2}, \frac{\sqrt{3}r}{2})\) and \((\frac{r}{2}, \frac{\sqrt{3}r}{2})\).
Finding the Midpoint of CD
Let M be the midpoint of CD. The coordinates of M are given by the midpoint formula:
\(M = \left(\frac{x_C + x_D}{2}, \frac{y_0 + y_0}{2}\right)\)
\(M = \left(\frac{-r/2 + r/2}{2}, \frac{\sqrt{3}r/2 + \sqrt{3}r/2}{2}\right)\)
\(M = \left(\frac{0}{2}, \frac{2(\sqrt{3}r/2)}{2}\right)\)
\(M = \left(0, \frac{\sqrt{3}r}{2}\right)\)
If we had chosen \(y_0 = -\frac{\sqrt{3}r}{2}\), the midpoint would be \((0, -\frac{\sqrt{3}r}{2})\).
Determining the Locus of the Midpoint
The quadrilateral ABCD rotates about the center O(0, 0). When the quadrilateral rotates, the distance of any point on the figure from the center O remains constant. We are interested in the locus of the midpoint M of CD.
The distance of the midpoint M \((0, \frac{\sqrt{3}r}{2})\) from the center O \((0, 0)\) is calculated using the distance formula:
\(Distance(O, M) = \sqrt{(0-0)^2 + (\frac{\sqrt{3}r}{2} - 0)^2}\)
\(Distance(O, M) = \sqrt{0^2 + (\frac{\sqrt{3}r}{2})^2}\)
\(Distance(O, M) = \sqrt{\frac{3r^2}{4}}\)
\(Distance(O, M) = \frac{\sqrt{3}r}{2}\)
This distance is constant regardless of the rotation angle, because the geometry of the chord CD relative to the center O (its length and distance from O) is fixed. The locus of a point that maintains a constant distance from a fixed center is a circle.
Therefore, the locus of the middle point of CD is a circle centered at O with a radius equal to the distance of M from O.
The radius of the locus circle is \(\frac{\sqrt{3}r}{2}\).
Summary of Steps
- Established the coordinate system with O at origin and AB on the x-axis.
- Used the properties of the chord CD (parallel to diameter AB, length \(r\)) to find its possible y-coordinates.
- Calculated the coordinates of the midpoint M of CD.
- Determined the distance of M from the center O.
- Concluded that the locus is a circle with this constant distance as its radius.
The radius of the locus of the middle point of CD is \(\frac{\sqrt{3}r}{2}\).
Revision Table: Concyclic Quadrilateral and Circle Properties
| Concept | Description | Relevance to Problem |
|---|---|---|
| Concyclic Quadrilateral | A quadrilateral whose vertices all lie on a single circle. | ABCD is concyclic, ensuring A, B, C, D are on the given circle. |
| Diameter | A chord passing through the center of the circle. Longest chord. | AB is the diameter (length \(2r\)), fixing its position through O. |
| Chord Parallel to Diameter | A chord that is horizontally or vertically aligned if the diameter is vertical or horizontal, respectively. | CD is parallel to AB, simplifying its coordinate representation (constant y-coordinate). |
| Locus | The set of all points satisfying a given condition. | We found the set of all possible positions of the midpoint of CD during rotation. |
| Rotation about Center | Movement where every point maintains its distance from the center of rotation. | Ensures the distance of M from O remains constant, defining the locus as a circle. |
Additional Information: Chords and Their Midpoints
When considering chords in a circle centered at the origin \(x^2 + y^2 = r^2\):
- A chord parallel to the x-axis has a constant y-coordinate, say \(y_0\). Its endpoints are \((\pm \sqrt{r^2 - y_0^2}, y_0)\).
- The length of such a chord is \(2\sqrt{r^2 - y_0^2}\).
- The midpoint of such a chord is \((0, y_0)\). The distance of the midpoint from the center (0,0) is \(|y_0|\).
- In this problem, we worked backwards: we knew the length of the chord CD (\(r\)) and its orientation (parallel to AB, i.e., horizontal). We used the length formula \(r = 2\sqrt{r^2 - y_0^2}\) to find the possible values of \(|y_0|\).
- \(r^2 = 4(r^2 - y_0^2)\)
- \(r^2 = 4r^2 - 4y_0^2\)
- \(4y_0^2 = 3r^2\)
- \(y_0^2 = \frac{3r^2}{4}\)
- \(|y_0| = \frac{\sqrt{3}r}{2}\)
This confirms that any horizontal chord of length \(r\) in a circle of radius \(r\) must be at a distance of \(\frac{\sqrt{3}r}{2}\) from the center. The midpoint of such a chord lies on the y-axis (if the diameter is on the x-axis) at this distance. As the circle rotates, this midpoint traces a circle with this distance as its radius.
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