Two Circles Touch Internally

If two given circles are touches internally with each other, take an example to understand the concept of internally touches circles.


circle-touches-internally

Consider the given circles
{x^2} + {y^2} + 2x - 8 = 0\,\,\,{\text{ -  -  - }}\left( {\text{i}} \right)and {x^2} + {y^2} - 6x + 6y - 46 = 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 centre and radius, we have
Centre {C_1}\left( { - g, - f} \right) = {C_1}\left( { - 1,0} \right) and
Radius {r_1} = \sqrt {{g^2} + {f^2} - c}  = \sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2} - \left( { - 8} \right)}  = \sqrt {1 + 8}  = \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 centre and radius, we have
Centre {C_2}\left( { - g, - f} \right) = {C_2}\left( { - \left( { - 3} \right), - 3} \right) = {C_2}\left( {3, - 3} \right) and
Radius {r_2} = \sqrt {{g^2} + {f^2} - c}  = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 3} \right)}^2} - \left( { - 46} \right)}  = \sqrt {9 + 9 + 46}  = \sqrt {64}  = 8
Using the distance formula, we find the distance between the centres of the given circles, we have

\left| {{C_1}{C_2}} \right| = \sqrt {{{\left( {3 - \left( { - 1} \right)} \right)}^2} + {{\left( { - 3 - 0} \right)}^2}}  = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( { - 3} \right)}^2}}  = \sqrt {16 + 9}  = \sqrt {25}  = 5


Now subtracting the radius from second from the first circle, we have

{r_2} - {r_1} = 8 - 3 = 5


This shows that the distance between the centres of given circles is equal to the difference of their radii. This is only possible if the circle touches each other internally as shown in the given diagram.

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