Two circles {x^2} + {y^2} - 2x - 3 = 0 and {x^2} + {y^2} - 4x - 6y - 8 = 0 are such that

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10-1.Circle and System of Circles

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The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that

A

They touch each other

B

They intersect each other

C

One lies inside the other

D

None of these

Solution

(b) ${r_1} + {r_2} = 2 + \sqrt {21} $,

${C_1}{C_2} $ $= \sqrt {{1^2} + {3^2}} = \sqrt {10} $

Obviously ${r_1} + {r_2} > {C_1}{C_2}$.

Hence circles cut each other.

Standard 11

Mathematics

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