# 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

$\left| {{C_1}{C_2}} \right| = \sqrt {{{\left( {3 – \left( { – 1} \right)} \right)}^2} + {{\left( { – 2 – 1} \right)}^2}} = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( { – 3} \right)}^2}} = \sqrt {16 + 9} = \sqrt {25} = 5$

${r_1} + {r_2} = 3 + 2 = 5$