Interaction of Two Circles

When two circles are drawn close to each other, several interesting things can happen. They might cut each other at two points, touch each other at exactly one point, or not meet at all. In this section, we will study how two circles interact depending on how far apart they are and how big they are.

We will learn how to find out whether they intersect, and if they do, how they meet—at what angle, and whether they are orthogonal (i.e., meet at right angles). We will also look at important ideas like the common chord if they intersect, or common tangents if they don't. We will define and use the concept of the radical axis.

Relative Location of Two Circles

To understand how two circles behave when placed in a plane, the first and most natural question is: how are they positioned relative to each other?

Let us consider two circles:

\[ S_1: (x - x_1)^2 + (y - y_1)^2 = r_1^2 \quad \text{and} \quad S_2: (x - x_2)^2 + (y - y_2)^2 = r_2^2 \]

Here, \( C_1(x_1, y_1) \) and \( C_2(x_2, y_2) \) are the centres of the circles, and \( r_1, r_2 \) are their respective radii. The distance between the centres is denoted by

\[ d = C_1C_2 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

This distance \( d \), together with the sizes \( r_1 \) and \( r_2 \), determines the mutual position of the two circles. There are five possibilities:

Case I: The circles are externally apart

When the distance between the centres is more than the sum of the radii:

\[ d > r_1 + r_2 \]

The circles are far from each other and do not touch or intersect.

Case II: The circles touch each other externally

When the circles are brought closer so that they just touch at one point externally, then

\[ d = r_1 + r_2 \]

The point of contact lies on the line joining the centres.

Case III: The circles intersect at two distinct points

In this case, the circles overlap partially. This happens when:

\[ |r_1 - r_2| < d < r_1 + r_2 \]

The condition \( d < r_1 + r_2 \) ensures that the circles are close enough to interact, but we must also have \( d > |r_1 - r_2| \), else one circle may lie completely inside the other or touch it internally. The need for this second inequality becomes clearer when compared with the next two cases.

We can interpret this case from a different perspective—using triangle geometry.

Suppose the circles \( S_1 \) and \( S_2 \) intersect at two distinct points. Let \( A \) be one such point of intersection. Then the triangle \( \triangle C_1AC_2 \) is formed, where \( C_1A = r_1 \) and \( C_2A = r_2 \). The side \( C_1C_2 \) is the distance between the centres, denoted by \( d \).

Since \( \triangle C_1AC_2 \) is a genuine triangle (i.e., not degenerate), the triangle inequality must hold for the three sides \( r_1, r_2, d \):

\[ \begin{align} r_1 + r_2 &> d \tag{1} \\ r_1 + d &> r_2 \tag{2} \\ r_2 + d &> r_1 \tag{3} \end{align} \]

From inequalities (2) and (3), we get:

\[ \begin{align} d &> r_2 - r_1 \tag{4} \\ d &> r_1 - r_2 \tag{5} \end{align} \]

Taken together, these imply:

\[ d > |r_1 - r_2| \tag{6} \]

Combining this with inequality (1), we conclude that the necessary and sufficient condition for two circles to intersect at two distinct points is:

\[ |r_1 - r_2| < d < r_1 + r_2 \]

Case IV: The circles touch each other internally

Suppose without loss of generality that \( r_1 > r_2 \). Then the smaller circle lies entirely within the larger and they touch at exactly one point. The centres lie on the same line through the point of contact, and

\[ d = r_1 - r_2 \quad \text{or more generally,} \quad d = |r_1 - r_2| \]

This is a limiting case just like Case II, but from the inside.

Case V: One circle lies completely inside the other

Now the circles do not touch and one is entirely inside the other. The distance between the centres is strictly less than the difference of the radii:

\[ d < |r_1 - r_2| \]

From these observations, we note that the condition \( d < r_1 + r_2 \) alone is not enough to ensure that the circles intersect. It is satisfied in Cases III, IV, and V, but only in Case III do the circles actually intersect at two distinct points. Therefore, to ensure that circles cut each other,

\[ |r_1 - r_2| < d < r_1 + r_2 \]

must hold.

Angle of Intersection of Two Circles

When two circles intersect each other at two points, one natural question arises: at what angle do they intersect?

To recall, the angle of intersection of two curves at a common point is defined as the angle between their tangents drawn at that point. In general, for arbitrary curves, one standard method to compute this angle uses calculus: find the point of intersection, differentiate both curves to obtain the slopes \( m_1 \) and \( m_2 \) of the tangents at that point, and then apply the angle formula

\[ \tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

However, in the case of circles, this method becomes unnecessarily tedious. The calculation of intersection points and then the implicit differentiation of each circle, followed by finding slopes at points of intersection, introduces avoidable complexity.

Instead, we shall use a far more elegant and geometrical method for circles.

Suppose two circles \( S_1 \) and \( S_2 \) with centres \( C_1 \) and \( C_2 \), and radii \( r_1 \) and \( r_2 \), intersect at two distinct points \( A \) and \( B \). Let the distance between their centres be denoted by \( d = C_1C_2 \). Due to symmetry, the angle of intersection is the same at both points \( A \) and \( B \). We focus on point \( A \).

Join the radii \( C_1A \) and \( C_2A \). At point \( A \), draw tangents to both circles, say \( AT_1 \) and \( AT_2 \), such that \( AT_1 \) is tangent to \( S_1 \) and \( AT_2 \) is tangent to \( S_2 \). Let \( \theta \) be the angle between these two tangents. Then, by the geometry of a circle, each tangent is perpendicular to the radius drawn to the point of contact. Hence,

\[ \angle C_1AT_1 = 90^\circ, \quad \angle C_2AT_2 = 90^\circ \]

Then,

\[ \angle C_1AT_2 = 90^\circ - \theta, \quad \angle C_2AT_1 = 90^\circ - \theta \]

Since \( \angle T_1AT_2 = \theta \), it follows that:

\[ \angle C_1AC_2 = \angle T_2AC_1 + \angle T_1AT_2 + \angle T_1AC_2 = (90^\circ - \theta) + \theta + (90^\circ - \theta) = 180^\circ - \theta \]

Therefore, in triangle \( \triangle C_1AC_2 \), we have:

\[ \angle C_1AC_2 = 180^\circ - \theta \]

Now, using the Law of Cosines in triangle \( \triangle C_1AC_2 \), with sides \( C_1A = r_1 \), \( C_2A = r_2 \), and \( C_1C_2 = d \), we have:

\[ \cos(180^\circ - \theta) = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \]

But \( \cos(180^\circ - \theta) = -\cos\theta \), so:

\[ -\cos\theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \quad \implies \quad \cos\theta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2} \]

Hence, the angle of intersection \( \theta \) between the two circles satisfies:

\[ \cos\theta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2} \]

Orthogonal Circles

When two circles intersect orthogonally, the angle of intersection \( \theta \) is \( 90^\circ \). From the previous formula for angle of intersection:

\[ \cos\theta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2} \]

orthogonality implies \( \theta = 90^\circ \Rightarrow \cos\theta = 0 \). Substituting:

\[ \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2} = 0 \implies d^2 = r_1^2 + r_2^2 \]

This gives a necessary and sufficient condition for orthogonality of intersecting circles in terms of their centres and radii.

We can also see this geometrically. Let the circles \( S_1 \) and \( S_2 \) intersect orthogonally at point \( A \). The tangent to \( S_1 \) at \( A \), say \( AT_1 \), is perpendicular to radius \( C_1A \). But since the circles intersect orthogonally, it can be seen that this tangent is also normal to \( S_2 \). Thus, \( AT_1 \) must pass through \( C_2 \). By the same logic, the tangent to \( S_2 \) at \( A \) passes through \( C_1 \). Therefore, triangle \( \triangle C_1AC_2 \) is right-angled at \( A \). By Pythagoras:

\[ C_1C_2^2 = C_1A^2 + C_2A^2 \implies d^2 = r_1^2 + r_2^2 \]

Now, suppose the equations of the circles are:

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

Then the centres are:

\[ C_1 = (-g_1, -f_1), \quad C_2 = (-g_2, -f_2) \]

and the radii are:

\[ r_1 = \sqrt{g_1^2 + f_1^2 - c_1}, \quad r_2 = \sqrt{g_2^2 + f_2^2 - c_2} \]

The square of the distance between centres is:

\[ d^2 = (g_1 - g_2)^2 + (f_1 - f_2)^2 \]

Now impose the orthogonality condition \( d^2 = r_1^2 + r_2^2 \). Substituting:

\[ (g_1 - g_2)^2 + (f_1 - f_2)^2 = (g_1^2 + f_1^2 - c_1) + (g_2^2 + f_2^2 - c_2) \]

Exapanding:

\[ g_1^2 + g_2^2 - 2g_1g_2 + f_1^2 + f_2^2 - 2f_1f_2 = g_1^2 + f_1^2 - c_1 + g_2^2 + f_2^2 - c_2 \]

Canceling identical terms from both sides yields:

\[ -2g_1g_2 - 2f_1f_2 = -c_1 - c_2 \implies 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]

Thus, two circles intersect orthogonally if and only if:

\[ \boxed{2g_1g_2 + 2f_1f_2 = c_1 + c_2} \]

Example

Prove that any circle orthogonal to the circle \( x^2 + y^2 = r^2 \) must have the form

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

for some real numbers \( g, f \in \mathbb{R} \).

Solution.

Let the required circle be of the general form

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \tag{1} \]

which has centre \( (-g, -f) \) and radius \( \sqrt{g^2 + f^2 - c} \), assuming the expression under the square root is positive.

The given circle is

\[ x^2 + y^2 - r^2 = 0 \tag{2} \]

which has centre \( (0, 0) \), and radius \( r \). Thus, \( g_1 = f_1 = 0 \), and \( c_1 = -r^2 \).

Using the orthogonality condition between two circles:

\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]

Substituting the parameters from (2) and (1):

\[ 2(0)(g) + 2(0)(f) = -r^2 + c \implies 0 = -r^2 + c \implies c = r^2 \]

Substituting this value of \( c \) back into (1), the equation of the required circle becomes:

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

as required.

Common Chord of Two Intersecting Circles

Let two circles intersect each other at points \( A \) and \( B \). Their general equations are:

\[ \begin{aligned} S_1 &: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 \\ S_2 &: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 \end{aligned} \]

Since both circles pass through points \( A \) and \( B \), the chord \( AB \) lies on both circles. We define this line segment as the common chord.

We claim that the equation of the common chord is given by:

\[ S_1 - S_2 = 0 \]

Computing this difference:

\[ \begin{aligned} S_1 - S_2 &= (x^2 + y^2 + 2g_1x + 2f_1y + c_1) - (x^2 + y^2 + 2g_2x + 2f_2y + c_2) \\ &= 2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) \end{aligned} \]

Thus, the equation of the common chord is:

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

This is a linear equation and hence represents a straight line.

To confirm that this is indeed the line \( AB \), observe the following:

1. The equation \( S_1 - S_2 = 0 \) is a linear equation and hence represents a straight line.
2. Any point \( P \) lying on both circles must satisfy both \( S_1 = 0 \) and \( S_2 = 0 \). Subtracting, we find \( S_1(P) - S_2(P) = 0 \), so \( P \) satisfies the equation \( S_1 - S_2 = 0 \).
3. In particular, points \( A \) and \( B \), the points of intersection, satisfy both \( S_1 = 0 \) and \( S_2 = 0 \), hence satisfy \( S_1 - S_2 = 0 \).
4. Only one straight line passes through two distinct points \( A \) and \( B \). Therefore, the line \( S_1 - S_2 = 0 \) is precisely the common chord \( AB \).

Length of the Common Chord of Two Intersecting Circles

Let two circles intersect at points \( A \) and \( B \). The segment \( AB \), which lies on both circles, is called the common chord. To compute its length, we use the standard method of finding the length of a chord of a circle.

Suppose the two circles are:

\[ \begin{aligned} S_1 &: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 \\ S_2 &: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 \end{aligned} \]

Then the equation of the common chord is obtained by subtracting:

\[ S_1 - S_2 = 0 \Rightarrow 2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = 0 \tag{1} \]

Let us use circle \( S_1 \) to compute the length of the chord. Its centre is

\[ C_1 = (-g_1, -f_1) \]

and its radius is

\[ r = \sqrt{g_1^2 + f_1^2 - c_1} \]

The length of a chord in a circle at a perpendicular distance \( d \) from the centre is given by:

\[ \text{Length} = 2\sqrt{r^2 - d^2} \]

To compute \( d \), the perpendicular distance from the centre \( C_1 \) to the chord (1), we use the formula for distance from a point to a line:

\[ d = \frac{\left|2(g_1 - g_2)(-g_1) + 2(f_1 - f_2)(-f_1) + (c_1 - c_2)\right|}{\sqrt{[2(g_1 - g_2)]^2 + [2(f_1 - f_2)]^2}} \]

Now substituting this \( d \) and \( r \) into the chord length formula, we get the full expression for the length of the common chord. Hence, the length is:

\[ \boxed{\text{Length of common chord} = 2\sqrt{r^2 - d^2}} \]

where:

• \( r = \sqrt{g_1^2 + f_1^2 - c_1} \)
• \( d = \) perpendicular distance from \( (-g_1, -f_1) \) to the line \( S_1 - S_2 = 0 \)

The same formula applies if we choose the second circle instead, since both circles intersect at the same chord.

Length of Common Chord — Geometrical Method

Suppose two circles \( S_1 \) and \( S_2 \) intersect at points \( A \) and \( B \), with radii \( r_1 \) and \( r_2 \), and centres \( C_1 \) and \( C_2 \) respectively. Let the distance between the centres be \( d = C_1C_2 \). The common chord \( AB \) is bisected at right angles by the line joining the centres, so let \( N \) be the midpoint of \( AB \), lying on both \( C_1C_2 \) and the chord.

By symmetry, \( AN = NB \), so

\[ AB = 2AN \]

Now \( AN \) is the altitude from point \( A \) to base \( C_1C_2 \) in triangle \( \triangle C_1AC_2 \). Hence, the area of triangle \( \triangle C_1AC_2 \) is:

\[ \text{Area} = \frac{1}{2} \cdot C_1C_2 \cdot AN = \frac{1}{2} \cdot d \cdot AN \implies AN = \frac{2(\text{Area of } \triangle C_1AC_2)}{d} \]

Therefore,

\[ AB = 2AN = \frac{4(\text{Area of } \triangle C_1AC_2)}{d} \]

We now compute the area of triangle \( C_1AC_2 \) using Heron’s formula. The side lengths are:

\[ a = r_1, \quad b = r_2, \quad c = d \]

Thus, the semi-perimeter is:

\[ s = \frac{r_1 + r_2 + d}{2} \]

Compute the parts of Heron’s formula:

\[ \begin{aligned} s - r_1 &= \frac{r_2 - r_1 + d}{2}, \\ s - r_2 &= \frac{r_1 - r_2 + d}{2}, \\ s - d &= \frac{r_1 + r_2 - d}{2} \end{aligned} \]

Then:

\[ s(s - d) = \frac{(r_1 + r_2 + d)}{2} \cdot \frac{(r_1 + r_2 - d)}{2} = \frac{(r_1 + r_2)^2 - d^2}{4} \]

\[ (s - r_1)(s - r_2) = \frac{(r_2 - r_1 + d)}{2} \cdot \frac{(r_1 - r_2 + d)}{2} = \frac{d^2 - (r_1 - r_2)^2}{4} \]

Hence, the area of triangle \( C_1AC_2 \) is:

\[ \text{Area} = \sqrt{s(s - r_1)(s - r_2)(s - d)} = \frac{1}{4} \sqrt{[(r_1 + r_2)^2 - d^2][d^2 - (r_1 - r_2)^2]} \]

Therefore,

\[ AB = \frac{4 \cdot \text{Area}}{d} = \frac{\sqrt{[(r_1 + r_2)^2 - d^2][d^2 - (r_1 - r_2)^2]}}{d} \]

\[ \boxed{AB = \frac{\sqrt{[(r_1 + r_2)^2 - d^2][d^2 - (r_1 - r_2)^2]}}{d}} \]

This expression gives the length of the common chord in terms of the radii and the distance between the centres.

Special Cases: Length of the Common Chord

We now consider two important special cases.

Case 1: Equal Radii \( r_1 = r_2 = r \)

Substituting into the general formula:

\[ AB = \frac{\sqrt{[(2r)^2 - d^2][d^2 - 0^2]}}{d} = \frac{\sqrt{(4r^2 - d^2)d^2}}{d} = \sqrt{4r^2 - d^2} \]

\[ \boxed{AB = \sqrt{4r^2 - d^2}} \]

Case 2: Circles Intersect Orthogonally

In this case, the angle \( \angle C_1AC_2 = 90^\circ \). Then the area of triangle \( C_1AC_2 \) is:

\[ \text{Area} = \frac{1}{2} r_1 r_2 \]

Hence, the length of the common chord is:

\[ AB = 2AN = \frac{4(\text{Area})}{d} = \frac{4 \cdot \frac{1}{2} r_1 r_2}{d} = \frac{2r_1r_2}{d} \]

\[ \boxed{AB = \frac{2r_1r_2}{d}} \]

A Circle Bisecting Another Circle

By this, we mean that a circle \( S_1 \) bisects the circumference of another circle \( S_2 \). That is, the two circles intersect each other at two points \( A \) and \( B \), and the common chord \( AB \) divides the circumference of \( S_2 \) into two equal arcs.

This condition implies that the chord \( AB \), which is common to both circles, is a diameter of circle \( S_2 \). Hence, the common chord passes through the centre of \( S_2 \).

Let:

• \( C_1 \), \( C_2 \) be the centres of \( S_1 \), \( S_2 \),
• \( r_1, r_2 \) be their respective radii,
• \( d = C_1C_2 \) be the distance between the centres.

Since \( AB \) is a chord of \( S_1 \), the perpendicular dropped from its centre \( C_1 \) to the chord will bisect it. And since \( AB \) is also a diameter of \( S_2 \), the centre of \( S_2 \), which is \( C_2 \), lies on the chord and is its midpoint. Thus, the foot of the perpendicular from \( C_1 \) to chord \( AB \) is \( C_2 \).

Hence, triangle \( \triangle C_1AC_2 \) is a right-angled triangle with the right angle at \( C_2 \).

Therefore, by the Pythagorean Theorem:

\[ C_1A^2 = C_2A^2 + C_1C_2^2 \implies r_1^2 = r_2^2 + d^2 \]

\[ \boxed{r_1^2 = r_2^2 + d^2} \]

This condition characterizes the case when one circle bisects another.

Limiting Case of Common Chord

When two circles intersect at two distinct points \( A \) and \( B \), the line segment \( AB \) is the common chord, and its equation is given by:

\[ S_1 - S_2 = 0 \]

As the two circles gradually move apart, the points of intersection \( A \) and \( B \) begin to approach each other. The common chord \( AB \) becomes shorter and shorter. In the limiting case, when \( A \) and \( B \) coalesce into a single point \( P \), the circles touch each other at exactly one point.

In this scenario, the common chord degenerates into a common tangent at the point of contact \( P \). The equation

\[ S_1 - S_2 = 0 \]

remains valid and now represents this common tangent. Thus, the common chord becomes the common tangent in the limiting case when the two points of intersection merge into one.

Hence, as the length of the common chord tends to zero, it transforms into the common tangent at the point where the two circles touch each other.