Circles {x^2} + {y^2} - 2x - 4y = 0 and {x^2} + {y^2} - 8y - 4 = 0

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10-1.Circle and System of Circles

medium

Circles ${x^2} + {y^2} - 2x - 4y = 0$ and ${x^2} + {y^2} - 8y - 4 = 0$

A

Touch each other internally

B

Touch each other externally

C

Cuts each other at two points

D

None of these

(IIT-1973)

Solution

(a) ${C_1}(1,\;2),\;{C_2}(0,\;4),\;{R_1} = \sqrt 5 ,\;{R_2} = 2\sqrt 5 $

${C_1}{C_2} = \sqrt 5 $ and ${C_1}{C_2} = \;|{R_2} – {R_1}|$

Hence circles touch internally.

Standard 11

Mathematics

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