Radical Axis

The radical axis of two non-concentric circles is the set of points that have the same power with respect to both circles.
example of radical axis

The power of a point \( P \) with coordinates \((x_0, y_0)\) relative to a circle with center \((h, k)\) and radius \( r \) is given by the square of the distance from the point to the center of the circle minus the square of the circle's radius.

$$ \text{Power} = d^2 - r^2 $$

Where \( d \) is the distance from point \( P \) to the center of the circle, and \( r \) is the radius of the circle.

The distance between point \( P \) and the center of the circle \( (h, k) \) is given by:

$$ d = \sqrt{(x_0 - h)^2 + (y_0 - k)^2} $$

Therefore, the power of point \( P \) with respect to a circle in standard form is:

$$ \text{Power} = (x_0 - h)^2 + (y_0 - k)^2 - r^2 $$

The radical axis is the set of points where the difference in power with respect to the two circles is zero:

$$ (x_0 - h_1)^2 + (y_0 - k_1)^2 - r_1^2 = (x_0 - h_2)^2 + (y_0 - k_2)^2 - r_2^2$$

If the circles have general form equations, you can find the radical axis by subtracting one equation from the other. $$ x^2 - y^2 + ax + by + c = x^2 - y^2 + a'x + b'y + c' $$ $$ \require{cancel} \cancel{x^2} - \cancel{y^2} + ax + by + c = \cancel{x^2} - \cancel{y^2} + a'x + b'y + c' $$ $$ ax + by + c - a'x - b'y - c' = 0 $$ $$ (a - a')x + (b - b')y + c - c' = 0 $$ This results in a linear equation where x and y are the coordinates of points on the radical axis. I derive the variable y: $$ y = - \frac{a-a'}{b-b'} x - (c-c') $$  This allows me to obtain the slope of the radial axis: $$ m = - \frac{a-a'}{b-b'} $$

The radical axis is a straight line perpendicular to the line connecting the centers of the two circles.

example of a radical axis perpendicular to the line connecting the centers

For any point P on the radical axis, the lengths of the tangent segments drawn to the two circles are equal. For example, PC ≅ PF

example

If the two circles intersect, the radical axis passes through the two intersection points.

example of intersecting circles

If the two circles are tangent, the radical axis passes through the point of tangency.

example of tangent circles

    A Practical Example

    Consider two circles with the following standard form equations:

    $$ C_1 : \ (x - 4)^2 + (y - 4)^2 = 9 $$

    $$ C_2 : \ (x - 1)^2 + (y - 1)^2 = 4 $$

    The first circle has its center at point (4,4) and a radius of 3 since \( r^2 = 9 \) thus \( r = 3 \).

    The second circle has its center at point (1,1) and a radius of 2 since \( r^2 = 4 \) thus \( r = 2 \).

    the two circles

    The power of a general point P(x,y) with respect to the first circle centered at (h,k)=(4,4) with radius r=3 is:

    $$ W_1 = (x - h)^2 + (y - k)^2 - r^2 $$

    $$ W_1 = (x - 4)^2 + (y - 4)^2 - 3^2 $$

    $$ W_1 = (x - 4)^2 + (y - 4)^2 - 9 $$

    The power of the same point P(x,y) with respect to the second circle centered at (h,k)=(1,1) with radius r=2 is:

    $$ W_2 = (x - h)^2 + (y - k)^2 - r^2 $$

    $$ W_2 = (x - 1)^2 + (y - 1)^2 - 2^2 $$

    $$ W_2 = (x - 1)^2 + (y - 1)^2 - 4 $$

    To find the radical axis, set the powers equal:

    $$ W_1 = W_2 $$

    $$ (x - 4)^2 + (y - 4)^2 - 9 = (x - 1)^2 + (y - 1)^2 - 4 $$

    Simplify this equation to find the line representing the radical axis:

    $$ x^2 - 8x + 16 + y^2 - 8y + 16 - 9 = x^2 - 2x + 1 + y^2 - 2y + 1 - 4 $$

    $$ -8x - 8y + 23 = -2x - 2y - 2 $$

    $$ -2y + 8y = -8x + 23 + 2 + 2x $$

    $$ 6y = -6x + 25 $$

    $$ y = \frac{-6x + 25}{6} $$

    $$ y = -x + \frac{25}{6} $$

    Since the two circles intersect, the radical axis is the line passing through the two intersection points.

    example of radical axis between intersecting circles

    Example 2

    Consider two circles with general form equations:

    $$ C_1 : x^2 + y^2 - 8x - 8y + 23 = 0 $$

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

    In this case, I don't know the centers and radii. To find them, I would need to calculate them.

    However, if my goal is to find the equation of the radical axis, I can obtain it by simply subtracting one general form equation from the other.

    $$ C_1 = C_2 $$

    $$ x^2 + y^2 - 8x - 8y + 23 = x^2 + y^2 - 2x - 2y - 2 $$

    $$ \require{cancel} \cancel{x^2} + \cancel{y^2} - 8x - 8y + 23 = \cancel{x^2} + \cancel{y^2} - 2x - 2y - 2 $$

    This gives a linear equation:

    $$ -8y + 2y = 8x - 2x - 2 - 23 $$

    $$ -6y = 6x - 25 $$

    $$ 6y = -6x + 25 $$

    $$ y = \frac{-6x + 25}{6} $$

    $$ y = -x + \frac{25}{6} $$

    This is the equation of the radical axis of the two circles.

    example of radical axis between intersecting circles

    Example 3

    Consider two circles with general form equations:

    $$ C_1 : x^2 + y^2 - 2x = 0 $$

    $$ C_2 : x^2 + y^2 - 10x + 16 = 0 $$

    To obtain the equation of the radical axis, subtract the first from the second:

    $$ x^2 + y^2 - 2x = x^2 + y^2 - 10x + 16 $$

    $$ \cancel{x^2} + \cancel{y^2} - 2x = \cancel{x^2} + \cancel{y^2} - 10x + 16 $$

    $$ -2x = -10x + 16 $$

    $$ -2x + 10x = 16 $$

    $$ 8x = 16 $$

    $$ x = \frac{16}{8} $$

    $$ x = 2 $$

    In this case, the radical axis is a line parallel to the y-axis that passes through \( x = 2 \).

    Since the two circles are tangent, the radical axis passes exactly through the point of tangency of the two circles.

    the radical axis

    Example 4

    Consider two circles:

    $$ C_1 : (x - 7)^2 + (y - 7)^2 = 9 $$

    $$ C_2 : (x - 3)^2 + (y - 2)^2 = 4 $$

    The first circle has its center at (h,k) = (7,7) and a radius of \( r = \sqrt{9} = 3 \).

    The second circle has its center at (h,k) = (3,2) and a radius of \( r = \sqrt{4} = 2 \).

    example of external circles

    The power of a point P(x,y) with respect to the first circle centered at (h,k)=(7,7) with radius r=3 is:

    $$ W_1 = (x - h)^2 + (y - k)^2 - r^2 $$

    $$ W_1 = (x - 7)^2 + (y - 7)^2 - 3^2 $$

    $$ W_1 = (x - 7)^2 + (y - 7)^2 - 9 $$

    The power of the same point P(x,y) with respect to the second circle centered at (h,k)=(3,2) with radius r=2 is:

    $$ W_2 = (x - h)^2 + (y - k)^2 - r^2 $$

    $$ W_2 = (x - 3)^2 + (y - 2)^2 - 2^2 $$

    $$ W_2 = (x - 3)^2 + (y - 2)^2 - 4 $$

    Set the powers equal to find the radical axis:

    $$ W_1 = W_2 $$

    $$ (x - 7)^2 + (y - 7)^2 - 9 = (x - 3)^2 + (y - 2)^2 - 4 $$

    Then perform the necessary algebraic calculations:

    $$ x^2 - 14x + 49 + y^2 - 14y + 49 - 9 = x^2 - 6x + 9 + y^2 - 4y + 4 - 4 $$

    $$ \require{cancel} \cancel{x^2} - 14x + 49 + \cancel{y^2} - 14y + 49 - 9 = \cancel{x^2} - 6x + 9 + \cancel{y^2} - 4y + 4 - 4 $$

    $$ -14x - 14y + 98 - 9 = -6x - 4y + 9 $$

    $$ -14y + 4y = -6x + 14x - 89 + 9 $$

    $$ -10y = 8x - 80 $$

    $$ 10y = -8x + 80 $$

    $$ y = \frac{-8x + 80}{10} $$

    $$ y = - \frac{8x}{10} + \frac{80}{10} $$

    $$ y = - \frac{4x}{5} + 8 $$

    This is the equation of the radical axis of the two circles.

    In this case, the two circles are external, so the radical axis passes through the external space of the circles.

    the radical axis

    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.

    FacebookTwitterLinkedinLinkedin
    knowledge base

    Circumference

    Theorems

    Similarity