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Binomial Theorem

Question
CBSEENMA11015067
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In each of the following, find the equation of circle with centre open parentheses 1 half comma space 1 fourth close parentheses and radius 1 over 12

Solution

Centre (h, k) left right arrow open parentheses 1 half comma space 1 fourth close parentheses rightwards double arrowspace space space space space space space space space space space space space space straight h equals 1 half comma space straight k space equals space 1 fourth space space Radius comma space space straight r equals 1 over 12
∴ Equation of circle is : WiredFaculty
or open parentheses straight x minus 1 half close parentheses squared plus open parentheses straight y minus 1 fourth close parentheses squared space equals space open parentheses 1 over 12 close parentheses squared space or space straight x squared plus straight y squared minus straight x minus 1 half straight y plus 1 fourth plus 1 over 16 equals 1 over 144
or  144 straight x squared plus 144 straight y squared minus 144 straight x minus 72 straight y plus 36 plus 9 equals 1
or  144 straight x squared plus 144 straight y squared minus 144 straight x minus 72 straight y plus 44 equals 0
or  36 straight x squared plus 36 straight y squared minus 36 straight x minus 18 straight y plus 11 equals 0
Hence, the equation of circle is: 36 straight x squared plus 36 straight y squared minus 36 straight x minus 18 straight y plus 11 equals 0

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