# Two Circles Touch Externally

To understand the concept that the two given circles are touches externally with each other, take an example and to show that two circles touches externally,

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 centre and radius of the circle (i) respectively, now to find centre and radius compare the equation of circle with general equation of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ to get center and radius, we have
Centre ${C_1}\left( { - g, - f} \right) = {C_1}\left( { - 1, - \left( { - 1} \right)} \right) = {C_1}\left( { - 1,1} \right)$ and
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 centre and radius of the circle (ii) respectively, now to find centre and radius compare the equation of circle with general equation of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ to get center and radius, we have
Center ${C_2}\left( { - g, - f} \right) = {C_2}\left( { - \left( { - 3} \right), - 2} \right) = {C_2}\left( {3, - 2} \right)$ and
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, we have