# Two Circles Touching Externally

To understand the concept of two given circles that are touching  each other externally, look at this example. Consider the given circles

${x^2} + {y^2} + 2x - 2y - 7 = 0\,\,\,{\text{ - - - }}\left( {\text{i}} \right)$ and ${x^2} + {y^2} - 6x + 4y + 9 = 0\,\,\,{\text{ - - - }}\left( {{\text{ii}}} \right)$

Let ${C_1}$ and ${r_1}$ be the center and radius of the circle (i) respectively. Now to find the center and radius compare the equation of a circle with the general equation of a circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$.

To get the center and radius, we have

Center ${C_1}\left( { - g, - f} \right) = {C_1}\left( { - 1, - \left( { - 1} \right)} \right) = {C_1}\left( { - 1,1} \right)$
Radius ${r_1} = \sqrt {{g^2} + {f^2} - c} = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2} - \left( { - 7} \right)} = \sqrt {1 + 1 + 7} = \sqrt 9 = 3$

Let ${C_2}$ and ${r_2}$ be the center and radius of the circle (ii) respectively, Now to find the center and radius compare the equation of a circle with the general equation of a circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$.

To get the center and radius, we have

Center ${C_2}\left( { - g, - f} \right) = {C_2}\left( { - \left( { - 3} \right), - 2} \right) = {C_2}\left( {3, - 2} \right)$
Radius ${r_2} = \sqrt {{g^2} + {f^2} - c} = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( 2 \right)}^2} - 9} = \sqrt {9 + 4 - 9} = \sqrt 4 = 2$

First we find the distance between the centers of the given circles by using the distance formula from the analytic geometry, and we have

Now adding the radius of both the given circles, we have

This shows that the distance between the centers of the given circles is equal to the sum of their radii. This is only possible if the circles touche each other externally, as shown in the figure.