Tangent Circles

Two circles are called tangent circles if they intersect at exactly one point.
the line perpendicular to the radius

This common point is known as the point of tangency or the point of contact.

There are two ways circles can be tangent:

  • Internally tangent circles
    Two circles are internally tangent if one circle is inside the other and they touch at a single point. This occurs when the center of one circle lies within the other.
    circles tangent internally
    If the radii of the two circles are \( r_1 \) and \( r_2 \) and the distance between their centers is \( d \), the condition for internal tangency is: \[ d = |r_1 - r_2| \]

    Calculating the distance between centers. To calculate the distance between the centers, use the Euclidean distance formula between two points on a plane: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the centers \( C_1 \) and \( C_2 \) in the Cartesian plane.

  • Externally tangent circles
    This happens when the center of each circle is outside the other. This means the point of tangency is located between the two circles, on the line connecting their centers.
    the line perpendicular to the radius
    For external tangency, the sum of the radii \( r_1 + r_2 \) must equal the distance (d) between the centers of the two circles: \[ d = r_1 + r_2 \]

In both cases, tangent circles have a common point T and share the same tangent line t.

How to find the point of tangency

The point of tangency \( T \) lies along the line that connects the two centers.

If the circles are externally tangent, the coordinates of \( T \) can be found using the ratio of the radii to the distance \( d \):

$$ T = \left( x_1 + \frac{r_1}{r_1 + r_2}(x_2 - x_1), y_1 + \frac{r_1}{r_1 + r_2}(y_2 - y_1) \right) $$

If the circles are internally tangent, the coordinates of \( T \) can be found using the inverse ratio of the radii to the distance \( d \):

$$ T = \left( x_1 + \frac{r_1}{r_1 - r_2}(x_2 - x_1), y_1 + \frac{r_1}{r_1 - r_2}(y_2 - y_1) \right) $$

Alternatively, you can set up a system of equations using the equations of the two circles: $$ \begin{cases} (x - x_1)^2 + (y - y_1)^2 = r_1^2 \\ \\ (x - x_2)^2 + (y - y_2)^2 = r_2^2 \end{cases} $$ or  $$ \begin{cases} x^2 + y^2 + ax + by + c = 0 \\ \\ x^2 + y^2 + a'x + b'y + c' = 0 \end{cases} $$ Then solve the system using the reduction method by subtracting the first equation from the second.  $$ \begin{cases} x^2 + y^2 + ax + by + c = 0 \\ \\ (a' - a)x + (b' - b)y + (c' - c) = 0 \end{cases} $$ The second equation is the radical axis of the two circles, which, in the case of tangent circles, coincides with the tangent line between the two circles. At this point, solve the system to find the x and y coordinates of the point of tangency.  

Once the point of tangency \( T \) is known, the tangent line can also be found using the tangent line formula for a circle.

The equation of the tangent line to the circle \( (x - x_0)^2 + (y - y_0)^2 = r^2 \) with center \( C(x_0, y_0) \) and radius \( r \) at the point \( T = (x_T, y_T) \) is:

$$ (x_T - x_0)(x - x_0) + (y_T - y_0)(y - y_0) = r^2 $$

A practical example

Consider two circles:

$$ x^2 + y^2 - 2x - 9 = 0 $$

$$ C_2: x^2 + y^2 + 4x - 2y - 35 = 0 $$

The center and radius of the first circle can be found using the coefficients a = -2, b = 0, and c = -9

$$ C_1 \left( -\frac{a}{2}, -\frac{b}{2} \right) = \left( -\frac{-2}{2}, -\frac{0}{2} \right) = (1, 0) $$

$$ r_1 = \sqrt{ \left( -\frac{a}{2} \right)^2 + \left( -\frac{b}{2} \right)^2 - c } = \sqrt{1 - 0 - (-9)} = \sqrt{10} $$

Using the same process, find the center and radius of the second circle. In this case, the coefficients are a = 4, b = -2, and c = -35

$$ C_2 \left( -\frac{a}{2}, -\frac{b}{2} \right) = \left( -\frac{4}{2}, -\frac{-2}{2} \right) = (-2, 1) $$

$$ r_2 = \sqrt{ \left( -\frac{a}{2} \right)^2 + \left( -\frac{b}{2} \right)^2 - c } = \sqrt{(-2)^2 + (1)^2 - (-35)} = \sqrt{4 + 1 + 35} = \sqrt{40} $$

Once the coordinates of the centers \( C_1 (1, 0) \) and \( C_2 (-2, 1) \) are known, calculate the Euclidean distance between the two points

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

$$ d = \sqrt{(-2 - 1)^2 + (1 - 0)^2} $$

$$ d = \sqrt{(-3)^2 + (1)^2} $$

$$ d = \sqrt{9 + 1} $$

$$ d = \sqrt{10} $$

Calculate the sum of the radii and the absolute difference between the radii

$$ r_1 + r_2 = \sqrt{10} + \sqrt{40} = \sqrt{10} + \sqrt{4 \cdot 10} = \sqrt{10} + 2 \sqrt{10} = 3 \sqrt{10} $$

$$ | r_1 - r_2 | = | \sqrt{10} - \sqrt{40} | = | \sqrt{10} - 2 \sqrt{10} | = | -\sqrt{10} | = \sqrt{10} $$

In this case, the distance between the centers \( d = \sqrt{10} \) is equal to the absolute value of the difference between the radii

$$ d = | r _1 - r_2 | = \sqrt{10} $$

Therefore, the two circles are internally tangent.

The point of tangency is at the coordinates:

$$ T = \left( x_1 + \frac{r_1}{r_1 - r_2}(x_2 - x_1), y_1 + \frac{r_1}{r_1 - r_2}(y_2 - y_1) \right) $$

The radii of the two circles are \( r_1 = \sqrt{10} \) and \( r_2 = \sqrt{40} \)

$$ T = \left( x_1 + \frac{\sqrt{10}}{\sqrt{10} - \sqrt{40}}(x_2 - x_1), y_1 + \frac{\sqrt{10}}{\sqrt{10} - \sqrt{40}}(y_2 - y_1) \right) $$

The coordinates of the centers of the two circles are \( C_1(x_1, y_1) = (1, 0) \) and \( C_2(x_2, y_2) = (-2, 1) \)

$$ T = \left( 1 + \frac{\sqrt{10}}{\sqrt{10} - \sqrt{40}}(-2 - 1), 0 + \frac{\sqrt{10}}{\sqrt{10} - \sqrt{40}}(1 - 0) \right) $$

$$ T = \left( 1 + \frac{\sqrt{10}}{\sqrt{10} - 2\sqrt{10}}(-3), 0 + \frac{\sqrt{10}}{\sqrt{10} - 2\sqrt{10}}(1) \right) $$

$$ T = \left( 1 + \frac{\sqrt{10}}{-\sqrt{10}}(-3), 0 + \frac{\sqrt{10}}{\sqrt{10} - 2\sqrt{10}}(1) \right) $$

$$ T = \left( 1 + \frac{\sqrt{10}}{-\sqrt{10}}(-3), 0 + \frac{1}{1 - 2} \right) $$

$$ T = \left( 1 + 3, 0 - 1 \right) $$

$$ T = \left( 4, -1 \right) $$

Once the point of tangency \( T = (4, -1) \) is known, the equation of the tangent line can also be found.

$$ (x_T - x_0)(x - x_0) + (y_T - y_0)(y - y_0) = r^2 $$

$$ (4 - x_0)(x - x_0) + (-1 - y_0)(y - y_0) = r^2 $$

Consider the center and radius of one of the two circles, usually the simpler one, for example, \( C_1(1, 0) \) and \( r_1 = \sqrt{10} \)

$$ (4 - 1)(x - 1) + (-1 - 0)(y - 0) = (\sqrt{10})^2 $$

$$ 3(x - 1) - y = 10 $$

$$ 3x - 3 - y - 10 = 0 $$

$$ 3x - y - 13 = 0 $$

This is the equation of the tangent line passing through the point \( T(4, -1) \)

graph of tangent circles and point of tangency

 

Alternative Solution

Consider the two circles from the previous example:

$$ x^2 + y^2 - 2x - 9 = 0 $$

$$ x^2 + y^2 + 4x - 2y - 35 = 0 $$

In this case, set up a system of equations using the equations of the two circles:

$$ \begin{cases} x^2 + y^2 - 2x - 9 = 0 \\ \\ x^2 + y^2 + 4x - 2y - 35 = 0 \end{cases} $$

Apply the reduction method and subtract the first equation from the second (or vice versa).

$$ \begin{cases} x^2 + y^2 - 2x - 9 = 0 \\ \\ x^2 + y^2 + 4x - 2y - 35 - (x^2 + y^2 - 2x - 9) = 0 \end{cases} $$

$$ \begin{cases} x^2 + y^2 - 2x - 9 = 0 \\ \\ 6x - 2y - 26 = 0 \end{cases} $$

$$ \begin{cases} x^2 + y^2 - 2x - 9 = 0 \\ \\ 3x - y - 13 = 0 \end{cases} $$

The equation 3x - y - 13 = 0 is the radical axis, which coincides with the tangent line when the two circles are tangent.

Thus, the initial system is equivalent to the system formed by the equation of a circle and the equation of the radical axis.

$$ \begin{cases} x^2 + y^2 - 2x - 9 = 0 \\ \\ y = 3x - 13 \end{cases} $$

Substitute y in the first equation

$$ \begin{cases} x^2 + (3x - 13)^2 - 2x - 9 = 0 \\ \\ y = 3x - 13 \end{cases} $$

$$ \begin{cases} x^2 + (9x^2 - 2 \cdot 3x \cdot 13 + 13^2) - 2x - 9 = 0 \\ \\ y = 3x - 13 \end{cases} $$

$$ \begin{cases} x^2 + (9x^2 - 78x + 169) - 2x - 9 = 0 \\ \\ y = 3x - 13 \end{cases} $$

$$ \begin{cases} 10x^2 - 80x + 160 = 0 \\ \\ y = 3x - 13 \end{cases} $$

$$ \begin{cases} x^2 - 8x + 16 = 0 \\ \\ y = 3x - 13 \end{cases} $$

Now find the solutions of the quadratic equation.

$$ \Delta = b^2 - 4ac = (-8)^2 - 4 \cdot 16 = 64 - 64 = 0 $$

The quadratic equation has only one solution:

$$ x = \frac{-b \pm \sqrt{\Delta}}{2a} $$

$$ x = \frac{-(-8) \pm \sqrt{0}}{2 \cdot 1} $$

$$ x = \frac{8}{2} $$

$$ x = 4 $$

Once the x-coordinate is found, substitute it into the line equation to find the corresponding y-value.

$$ y = 3x - 13 $$

$$ y = 3 \cdot 4 - 13 $$

$$ y = 12 - 13 $$

$$ y = -1 $$

Therefore, the point of tangency between the two circles is at the point (x, y) = (4, -1)

$$ T(4, -1) $$

Here is the graph of the circles and the tangent line.

graph of tangent circles and point of tangency

Theorem of Tangent Circles

The necessary and sufficient condition for two circles to be tangent is:

  • The distance between the centers of the circles is equal to the absolute value of the difference of their radii if they are internally tangent: $$ OO'=| r-r' | $$
  • The distance between the centers of the circles is equal to the sum of their radii if they are externally tangent: $$ OO'=r+r' $$

Proof

Consider two tangent circles, the first with radius r and the second with radius r', assuming that r is greater than r'.

two tangent circles

Since the circles are tangent, they share a common point T, which is the point of tangency.

the two circles are tangent at point T

Point T is the endpoint of the radius for both circles:

$$ r = OT $$

$$ r' = O'T $$

At the point of tangency of a circle, there is a line t that is perpendicular to the radius.

Therefore, at point T, the same line is perpendicular to both radii r and r'.

the line perpendicular to the radius

As a result, points O, T, and O' are collinear.

Since these three points are collinear, there are two possible scenarios:

  • If the distance between the centers OO'=r+r' is equal to the sum of the radii, then the center of each circle lies outside the other, and the circles are externally tangent.
    the line perpendicular to the radius
  • If the distance between the centers OO'=r-r' is equal to the difference of the radii, then the center of one circle lies within the other, and the circles are internally tangent.
    the circles are internally tangent

Notes

Here are some observations about tangent circles:

  • When two circles are tangent, the point of tangency T lies on the segment OO' that connects the centers of the two circles.
    the line perpendicular to the radius

    Proof. Two tangent circles share a common point T (the point of tangency). At point T, a line "t" perpendicular to the radius passes through. Since there is only one line perpendicular to a segment, it follows that the tangent line "t" is perpendicular to both radii OT and O'T of the two circles. Consequently, the centers of the two circles O, O', and the point of tangency T are collinear, meaning they lie on the segment OO' and, therefore, on the same line.

And so on.

 

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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