Circle of Apollonius

The Circle of Apollonius is defined as the locus of a point \(P\) which moves in such a way that the ratio of its distances from two fixed points, \(A\) and \(B\), is a constant \(\lambda\), with \(\lambda \neq 1\). This geometric construct is named after the ancient Greek geometer Apollonius of Perga, who extensively studied these circles among other conic sections.

Mathematically, if \(PA\) is the distance from point \(P\) to point \(A\), and \(PB\) is the distance from point \(P\) to point \(B\), then the Circle of Apollonius is described by all points \(P\) satisfying the condition:

\[ \frac{PA}{PB} = \lambda, \]

where \(\lambda\) is a positive constant different from 1.

Geometrical Proof

Let \(A\) and \(B\) be fixed points on a plane, and let point \(P\) be a moving point such that the ratio \(AP:PB = \lambda\), where \(\lambda\) is a positive constant not equal to 1. Construct segment \(AB\) and consider an arbitrary position of \(P\). Construct \(PX\), the internal angle bisector of \(\angle APB\), and \(PY\), the external angle bisector of \(\angle APB\).

By the Angle Bisector Theorem, \(PX\) divides \(AB\) internally in the ratio \(AP:PB\), which is equal to \(\lambda\). Consequently, point \(X\) lies on segment \(AB\) and satisfies \(AX:XB = \lambda\). Similarly, \(PY\) divides the external segment extended from \(AB\) in the same ratio \(AP:PB\), which implies point \(Y\) lies on the extension of \(AB\) such that \(AY:YB = \lambda\).

Points \(X\) and \(Y\) are fixed since their positions are solely determined by the constant ratio \(\lambda\) and the fixed segment \(AB\). The angle \(\angle AYX\) is a straight angle, and the locus of point \(P\) must pass through both \(X\) and \(Y\) because these points satisfy the ratio \(AX:XB = \lambda\) and \(AY:YB = \lambda\), respectively.

Observe that the angle between line \(AX\) and \(AY\) is a right angle, thus indicating the locus of \(P\) is a circle with diameter \(XY\).

Radius ad Centre of Apollonius circle

Lat the center of the circle be \(C\) lying at the midpoint of \(XY\). By using the properties of the circle and the relationships established from the fixed points \(A\) and \(B\), we can determine the radius of the circle and the ratio of the segments \(AC\) and \(CB\).

First, to find the radius of the circle, we start by examining the segments \(AX\) and \(XB\). Since \(AX:XB = \lambda\), it follows that \(AX = \frac{\lambda}{\lambda + 1} \cdot AB\) and \(XB = \frac{1}{\lambda + 1} \cdot AB\).

Similarly, considering \(AY\) and \(YB\), with \(AY:YB = \lambda\), we obtain \(AY = \frac{\lambda}{1 - \lambda} \cdot AB\) and \(YB = \frac{1}{1 - \lambda} \cdot AB\).

The diameter \(XY\) of the circle can then be calculated as the difference between \(AY\) and \(AX\):

\[ XY = AY - AX = \left( \frac{\lambda}{1 - \lambda} - \frac{1}{\lambda + 1} \right) AB = \frac{2\lambda}{1 - \lambda^2} AB. \]

The radius of the circle, being half of the diameter, is thus:

\[ \text{Radius} = \frac{XY}{2} = \frac{\lambda}{1 - \lambda^2} AB. \]

Now, considering the center \(C\), we find \(AC\) by subtracting \(AX\) from \(CX\) (which is half of \(XY\)):

\[ AC = CX - AX = \frac{XY}{2} - AX = \frac{\lambda}{1 - \lambda^2} AB - \frac{\lambda}{\lambda + 1} AB = \frac{\lambda^2}{1 - \lambda^2} AB. \]

Since \(BC\) is simply the length of \(AB\) minus \(AC\), we have:

\[ BC = AB - AC = AB - \frac{\lambda^2}{1 - \lambda^2} AB = \frac{AB}{1 - \lambda^2}. \]

The ratio \(AC:CB\) thus becomes:

\[ AC:CB = \frac{\lambda^2}{1 - \lambda^2} AB : \frac{AB}{1 - \lambda^2} = \lambda^2. \]

The radius of the Circle of Apollonius is found to be:

\[ \text{Radius} = \left|\frac{\lambda}{1 - \lambda^2} \cdot AB\right|. \]

You need to take the modulus as in our proof \(\lambda<1\). If \(\lambda>1\), then we get radius equal to\(\frac{XY}{2} = \frac{\lambda}{ \lambda^2 - 1} AB\) Furthermore, the ratio in which center \(C\) divides the line segment \(AB\) externally is:

\[ AC:CB = \lambda^2. \]

Example

Given the fixed points \( A(-1,0) \) and \( B(3,0) \), with the ratio \( AP:PB = 2 \), we aim to determine the center and radius of the Apollonius circle for point \( P \).

Using the formula for the radius of the Apollonius circle:

\[ \text{Radius} = \left|\frac{\lambda}{1 - \lambda^2} \cdot AB\right|.\]

First, compute the distance \( AB \):

\[ AB = |3 - (-1)| = 4 \]

Given \( \lambda = 2 \), calculate the radius \( r \):

\[ r = \left|\frac{2}{1 - 2^2} \cdot 4\right| = \left|\frac{2}{1 - 4} \cdot 4 \right|= \frac{2}{3} \cdot 4 = \frac{8}{3} \]

To find the center \(C\) of the Apollonius circle given that \(A(-1,0)\) and \(B(3,0)\) are fixed points and the ratio \(AP:PB = 2\), we use the fact that \(C\) divides \(AB\) externally in the ratio \(\lambda^2 : 1\).

Given that \(\lambda = 2\), we have \(\lambda^2 = 4\). Now, we use the section formula for external division:

The coordinates of point \(C\) dividing \(AB\) externally in the ratio \(4:1\) are given by

\[ x_C = \frac{4x_B - x_A}{4 - 1} \quad \text{and} \quad y_C = \frac{4y_B - y_A}{4 - 1}. \]

Substituting the given coordinates of \(A\) and \(B\) into the formulas, we have

\[ x_C = \frac{4(3) - (-1)}{4 - 1} = \frac{12 + 1}{3} = \frac{13}{3}, \]

\[ y_C = \frac{4(0) - 0}{4 - 1} = 0. \]

Therefore, the center \(C\) is at \( \left(\frac{13}{3}, 0\right) \).

Hence, the center of the Apollonius circle is \(C\left(\frac{13}{3}, 0\right)\) and the radius is \(r = \frac{8}{3}\).