Externally Tangent Circles

Two circles are considered externally tangent if the distance between their centers, denoted as OO', exceeds the sum of their radii $$ \overline{OO'} > r+r' $$.

In this context, OO' indicates the distance between the centers of the two circles, and r and r' represent their respective radii.

Graphically, this means each circle does not overlap the other at any point.

External circles

When the distance OO' is greater than the sum of the radii r+r', each center, O and O', lies outside the circumference of the opposing circle.

Additionally, the addition of any center to its radius (O+r and O'+r') cannot intersect the other circle, ensuring they remain separate.

    A Practical Example

    Let's determine if the two circles defined by the following equations are externally tangent:

    $$ C_1 : x^2 + y^2 - 4x - 8y = -16 $$

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

    First, I'll rewrite the equations in standard form $ (x-h)^2 + (y-k)^2 = r^2 $ to find the center $ (h,k) $ and the radius $ r $ of each circle:

    • First equation: $$ x^2 + y^2 - 4x - 8y = -16 $$ Completing the square for \(x\) and \(y\) by adding and subtracting the necessary terms: $$ (x^2 - 4x + 4) + (y^2 - 8y + 16) = -16 + 4 + 16 $$ $$ (x - 2)^2 + (y - 4)^2 = 4 $$ Thus, the first circle has center \((2, 4)\) and radius \(r_1 = \sqrt{4} = 2\).
    • Second equation: $$ x^2 + y^2 - 2x + 4y = 2 $$ Completing the square for \(x\) and \(y\): $$ (x^2 - 2x + 1) + (y^2 + 4y + 4) = 2 + 1 + 4 $$ $$ (x - 1)^2 + (y + 2)^2 = 7 $$ Thus, the second circle has center \((1, -2)\) and radius \(r_2 = \sqrt{7} \approx 2.65\).

    Next, I calculate the distance between the centers of the circles using the Euclidean distance formula:

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

    The center of the first circle is \(O_1=(2, 4)\), so x1=2 and y1=4. The center of the second circle is \(O_2=(1, -2)\), so x2=1 and y2=-2.

    $$ d = \sqrt{(1 - 2)^2 + (-2 - 4)^2} $$

    $$ d = \sqrt{(-1)^2 + (-6)^2} $$

    $$ d = \sqrt{1 + 36} $$

    $$ d = \sqrt{37} \approx 6.08 $$

    Finally, I check the condition for the circles to be externally tangent.

    The circles are externally tangent if the distance between their centers is greater than the sum of their radii:

    $$ d > r_1 + r_2 $$

    $$ 6.08 > 2 + 2.65 $$

    $$ 6.08 > 4.65 $$

    Since \( 6.08 > 4.65 \), the two circles are externally tangent to each other.

    the circles are externally tangent

    Alternative Solution

    We need to determine if the two circles are externally tangent:

    $$ C_1 : x^2 + y^2 - 4x - 8y = -16 $$

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

    We use this formula to find the center of the circles $ ( - \frac{a}{2} , - \frac{b}{2} ) $ considering the circle equation in the form $ x^2 + y^2 + ax + by + c = 0 $:

    $$ C_1 : x^2 + y^2 - 4x - 8y + 16 = 0 $$

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

    For the first circle, the coefficients are a=-4 and b=-8.

    $$ O_1 = ( - \frac{a}{2} , - \frac{b}{2} ) = ( - \frac{-4}{2} , - \frac{-8}{2} ) = (2,4) $$

    For the second circle, the coefficients are a=-2 and b=4.

    $$ O_2 = ( - \frac{a}{2} , - \frac{b}{2} ) = ( - \frac{-2}{2} , - \frac{4}{2} ) = (1,-2) $$

    We use another formula to find the radii of the circles $ r = \sqrt{ ( - \frac{a}{2} )^2+ ( - \frac{b}{2} )^2 - c } $:

    For the first circle, a=-4, b=-8, c=16:

    $$ r_1 = \sqrt{ ( - \frac{a}{2} )^2+ ( - \frac{b}{2} )^2 - c } $$

    $$ r_1 = \sqrt{ 2^2 + 4^2 - 16 } $$

    $$ r_1 = \sqrt{ 4 + 16 - 16 } $$

    $$ r_1 = \sqrt{ 4 } $$

    $$ r_1 = 2 $$

    For the second circle, a=-2, b=4, c=-2:

    $$ r_2 = \sqrt{ ( - \frac{a}{2} )^2+ ( - \frac{b}{2} )^2 - c } $$

    $$ r_2 = \sqrt{ 1^2 + (-2)^2 - (-2) } $$

    $$ r_2 = \sqrt{ 1 + 4 + 2 } $$

    $$ r_2 = \sqrt{ 7 } $$

    $$ r_2 \approx 2.65 $$

    Next, I calculate the Euclidean distance between the centers of the two circles:

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

    The center of the first circle is \(O_1=(2, 4)\), so x1=2 and y1=4. The center of the second circle is \(O_2=(1, -2)\), so x2=1 and y2=-2.

    $$ d = \sqrt{(1 - 2)^2 + (-2 - 4)^2} $$

    $$ d = \sqrt{(-1)^2 + (-6)^2} $$

    $$ d = \sqrt{1 + 36} $$

    $$ d = \sqrt{37} \approx 6.08 $$

    Then, I check if the condition for external tangency is satisfied.

    Two circles are externally tangent if the distance between their centers is greater than the sum of their radii:

    $$ d > r_1 + r_2 $$

    $$ 6.08 > 2 + 2.65 $$

    $$ 6.08 > 4.65 $$

    Since \( 6.08 > 4.65 \), we can conclude that the two circles are externally tangent.

    And so forth.

     
     

    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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