Radical Axis

Radical Axis of Two Non-Concentric Circles

Let \( S_1 \) and \( S_2 \) be two given non-concentric circles with equations

\[ S_1 : x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0, \]

\[ S_2 : x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0. \]

Let a point \( P(x, y) \) move such that the lengths of the tangents drawn from \( P \) to both circles are equal. That is,

\[ \text{Length of tangent from } P \text{ to } S_1 = \text{Length of tangent from } P \text{ to } S_2. \]

From the known result on length of tangent from a point \( P(x, y) \) to a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \), the square of the length of the tangent is

\[ \text{Length}^2 = x^2 + y^2 + 2gx + 2fy + c = S(x, y). \]

So,

\[ PA^2 = S_1(x, y), \quad PB^2 = S_2(x, y), \]

and the condition \( PA = PB \) implies

\[ \sqrt{S_1} = \sqrt{S_2} \implies S_1 = S_2. \]

Therefore, the required locus is the set of all points satisfying

\[ S_1 - S_2 = 0. \]

Substituting the equations:

\[ (x^2 + y^2 + 2g_1x + 2f_1y + c_1) - (x^2 + y^2 + 2g_2x + 2f_2y + c_2) = 0, \]

\[ \implies 2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = 0. \]

This is the equation of a straight line and is called the radical axis of the two circles.
It has the important geometric property that it is the locus of all points from which tangents drawn to both circles are of equal length.

Given the two circles:

\[ S_1 : x^2 + y^2 + 4x + 6y - 1 = 0, \]

\[ S_2 : x^2 + y^2 + 3x - 2 = 0, \]

to find their radical axis, subtract the equations:

\[ S_1 - S_2 = (x^2 + y^2 + 4x + 6y - 1) - (x^2 + y^2 + 3x - 2) \]

\[ \implies (4x - 3x) + 6y + (-1 + 2) = 0 \]

\[ \implies x + 6y + 1 = 0 \]

Thus, the radical axis of the given pair of circles is the straight line

\[ \boxed{x + 6y + 1 = 0}. \]

Orientation of Radical Axis with Respect to the Two Circles

Let us consider two circles:

\[ S_1 : x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0, \]

\[ S_2 : x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0. \]

The radical axis is given by subtracting the two equations:

\[ S_1 - S_2 = 2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = 0. \]

This is the equation of a straight line. Its slope is given by:

\[ \text{slope of radical axis} = -\frac{2(g_1 - g_2)}{2(f_1 - f_2)} = -\frac{g_1 - g_2}{f_1 - f_2}. \]

Now, consider the centers of the two circles. The center of \( S_1 \) is \( (-g_1, -f_1) \), and the center of \( S_2 \) is \( (-g_2, -f_2) \). The slope of the line joining the centers is:

\[ \frac{-f_1 + f_2}{-g_1 + g_2} = \frac{f_2 - f_1}{g_2 - g_1} = \frac{f_1 - f_2}{g_1 - g_2}. \]

Hence, the product of slopes of the radical axis and the line joining the centers is:

\[ \left( -\frac{g_1 - g_2}{f_1 - f_2} \right) \cdot \left( \frac{f_1 - f_2}{g_1 - g_2} \right) = -1. \]

This confirms that the radical axis is perpendicular to the line joining the centers of the two circles.

\[ \text{Thus, the radical axis is always orthogonal to the line joining the centers of the given two circles.} \]

Position of Radical Axis When Two Circles Are Disjoint

Let two circles

\[ S_1 : x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0, \quad S_2 : x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 \]

be such that they are separated from each other—meaning they do not intersect and are not even touching.

The radical axis is defined as the locus of all points \( P(x, y) \) from which lengths of tangents drawn to both circles are equal. That is,

\[ \text{tangent length from } P \text{ to } S_1 = \text{tangent length from } P \text{ to } S_2. \]

Now assume the radical axis intersects one of the circles, say \( S_1 \). Let the point of intersection be \( P \). Since \( P \) lies on \( S_1 \), the tangent from \( P \) to \( S_1 \) has zero length:

\[ \text{tangent length to } S_1 = 0. \]

But \( P \) also lies on the radical axis, so the length of tangent to \( S_2 \) from \( P \) must also be zero. That implies \( P \) lies on \( S_2 \) as well. Hence, \( P \) is a common point of both circles, contradicting the assumption that the circles are disjoint.

Thus, we are forced to conclude:

• The radical axis cannot intersect either circle.
• Since it is a straight line perpendicular to the line joining the centers of the circles, and does not intersect either of them, it must lie entirely in between the two circles.

\[ \text{Therefore, when two circles are disjoint, the radical axis lies strictly between them and does not intersect either circle.} \]

When two circles intersect or touch each other, the radical axis takes on a very specific geometric form:

When two circles touch each other externally, the radical axis is the common tangent at their point of contact. This is immediately clear because the equation of the common tangent at the contact point is \( S_1 - S_2 = 0 \), which is precisely the equation of the radical axis.

When the two circles intersect at two distinct points, the radical axis is the common chord joining the points of intersection. This chord clearly lies on both circles and its equation again is \( S_1 - S_2 = 0 \), hence it is the radical axis.

When the two circles touch each other internally, the radical axis is once again the common tangent at the point of contact. The equation of this tangent is \( S_1 - S_2 = 0 \), and this shows that the radical axis and the common tangent are one and the same in this configuration too.

When one circle lies completely inside another without touching it, the two circles have no point in common, and all points on the smaller circle lie strictly within the interior of the larger one. In this case, the radical axis, defined by the equation \( S_1 - S_2 = 0 \), lies entirely outside both circles.

Another definition of Radical Axis

Another powerful and geometric way to understand the radical axis is through orthogonality. Let two circles be given, and suppose \( P \) is a point on their radical axis. From \( P \), draw tangents to the two circles, meeting them at points \( A \) and \( B \) respectively. Since \( P \) lies on the radical axis, the lengths of these tangents are equal, i.e., \( PA = PB \). Now, construct a circle centered at \( P \) with radius equal to this common tangent length. This new circle passes through both \( A \) and \( B \), and more importantly, it is orthogonal to each of the given circles: the tangents \( PA \) and \( PB \) from \( P \) are also radii of the new circle, and they meet the original circles at \( A \) and \( B \) perpendicularly. Thus, every point on the radical axis serves as the center of some circle orthogonal to both given circles. Hence, the radical axis can also be defined as the locus of centers of circles orthogonal to both given circles.

Let us now verify this definition algebraically. Let the two given circles be:

\[ S_1 : x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0, \]

\[ S_2 : x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0. \]

Let a third circle \( S \), with equation

\[ x^2 + y^2 + 2gx + 2fy + c = 0, \]

be orthogonal to both \( S_1 \) and \( S_2 \). Then the condition of orthogonality with each gives:

\[ 2gg_1 + 2ff_1 = c + c_1 \tag{1} \]

\[ 2gg_2 + 2ff_2 = c + c_2 \tag{2} \]

Subtracting (2) from (1), we obtain

\[ 2(g_1 - g_2)g + 2(f_1 - f_2)f = c_1 - c_2. \tag{3} \]

Now, since we are interested in the locus of the center of circle \( S \), let its center be \( (x, y) = (-g, -f) \). Substituting into (3),

\[ -2(g_1 - g_2)x - 2(f_1 - f_2)y = c_1 - c_2 \]

\[ \implies 2(g_1 - g_2)x + 2(f_1 - f_2)y + c_1 - c_2 = 0 \]

This is exactly the equation \( S_1 - S_2 = 0 \), which defines the radical axis. Hence, algebra confirms the geometric observation.

\[ \text{Thus, the radical axis is the locus of the centers of all circles orthogonal to two given circles.} \]

Radical Centre

Consider three circles \( S_1, S_2, S_3 \), whose centers are not collinear. Then each pair of circles determines a unique radical axis, which is a straight line defined by the locus of points from which the lengths of tangents to the two circles are equal.

Let the three radical axes be:

\[ R_1 : S_2 - S_3 = 0, \quad R_2 : S_3 - S_1 = 0, \quad R_3 : S_1 - S_2 = 0. \]

Observe that the equation of \( R_3 \) is simply the sum of the equations of \( R_1 \) and \( R_2 \). That is,

\[ R_1 + R_2 = (S_2 - S_3) + (S_3 - S_1) = S_2 - S_1 = R_3. \]

Hence, these three lines cannot be parallel or form a triangle—they are concurrent. The point of concurrency of these three radical axes is called the radical centre. When centres of circles are collinear, then all radical axes are parallel and hence they cannot intersect in one common point and there will be no radical centre.

At the radical centre, the point lies on all three radical axes, so the length of the tangents drawn from this point to all three circles is the same. That is,

\[ \text{Tangents from radical centre to } S_1, S_2, S_3 \text{ all have equal length}. \]

Furthermore, this point has a deeper geometric property: if we construct a circle centered at the radical centre with radius equal to the length of the tangent from this point to any of the three given circles, then this new circle will be orthogonal to all three circles. This follows from the fact that its radius meets each original circle tangentially at right angles, by construction.

Thus, the radical centre is not only the point of concurrency of the radical axes but also the unique point from which a single circle can be drawn orthogonal to all three given circles.

Example

Find the equation of the circle that is orthogonal to each of the following three circles:

\[ S_1 : x^2 + y^2 + x + y - 1 = 0, \quad S_2 : x^2 + y^2 + x - y - 3 = 0, \quad S_3 : x^2 + y^2 + 2x + 2y - 1 = 0. \]

Solution:

We begin by finding the radical centre, which is the point of intersection of the radical axes of each pair of circles.

Radical axis of \( S_1 \) and \( S_2 \):

Subtract the equations:

\[ (S_1 - S_2) \implies (x^2 + y^2 + x + y - 1) - (x^2 + y^2 + x - y - 3) = 2y + 2 = 0 \implies y = -1. \tag{1} \]

Radical axis of \( S_2 \) and \( S_3 \):

Subtract the equations:

\[ (S_2 - S_3) \implies (x^2 + y^2 + x - y - 3) - (x^2 + y^2 + 2x + 2y - 1) = -x - 3y - 2 = 0 \implies x + 3y + 2 = 0. \tag{2} \]

Substituting \( y = -1 \) from (1) into (2), we get:

\[ x + 3(-1) + 2 = 0 \implies x = 1. \]

Hence, the radical centre is the point \( P(1, -1) \).

To find the required circle, we construct a circle with centre at the radical centre and radius equal to the length of the tangent from this point to any of the given circles (all will give same length).

Using circle \( S_1 \), evaluate at \( (1, -1) \):

\[ S_1(1, -1) = 1^2 + (-1)^2 + 1 - 1 - 1 = 1. \]

Thus, the square of the tangent length is 1, and the radius is \( \sqrt{1} = 1 \).

Hence, the required circle has centre \( (1, -1) \) and radius \( 1 \), so the equation is:

\[ (x - 1)^2 + (y + 1)^2 = 1. \]