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CBSE Class 11-science Answered

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Asked by araima2001 | 27 Nov, 2016, 10:31: AM
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Given space equation space of space the space circle space left parenthesis straight S right parenthesis space is space straight x squared plus straight y squared minus 8 straight x minus 8 straight y space equals space 0 comma space it space passes space through space origin. Centre space equals space left parenthesis 4 comma space 4 right parenthesis space space and space radius space equals square root of 4 squared plus 4 squared end root equals 4 square root of 2 Given space equation space of space the space line space is space 7 straight x squared minus 18 xy plus 7 straight y squared equals 0 comma space it space also space passes space through space origin. space 7 straight x squared minus 18 xy plus 7 straight y squared equals 0 rightwards double arrow open parentheses square root of 7 straight x close parentheses squared minus 2 square root of 7 straight x cross times fraction numerator 9 over denominator square root of 7 end fraction straight y plus open parentheses square root of 7 straight y close parentheses squared plus open parentheses fraction numerator 9 over denominator square root of 7 end fraction straight y close parentheses squared minus open parentheses fraction numerator 9 over denominator square root of 7 end fraction straight y close parentheses squared equals 0 rightwards double arrow open parentheses square root of 7 straight x close parentheses squared minus 2 square root of 7 straight x cross times fraction numerator 9 over denominator square root of 7 end fraction straight y plus open parentheses fraction numerator 9 over denominator square root of 7 end fraction straight y close parentheses equals open parentheses fraction numerator 9 over denominator square root of 7 end fraction straight y close parentheses squared minus open parentheses square root of 7 straight y close parentheses squared rightwards double arrow open parentheses square root of 7 straight x minus fraction numerator 9 over denominator square root of 7 end fraction straight y close parentheses squared equals 81 over 7 straight y squared minus 7 straight y squared square root of 7 straight x minus fraction numerator 9 over denominator square root of 7 end fraction straight y equals plus-or-minus fraction numerator 4 square root of 2 over denominator square root of 7 end fraction straight y So comma space equation space of space individual space lines space space passing space through space origin space are square root of 7 straight x minus open parentheses fraction numerator 9 plus 4 square root of 2 over denominator square root of 7 end fraction close parentheses straight y equals 0................. open parentheses straight L subscript 1 close parentheses and space square root of 7 straight x minus open parentheses fraction numerator 9 minus 4 square root of 2 over denominator square root of 7 end fraction close parentheses straight y equals 0................. open parentheses straight L subscript 2 close parentheses According space to space the space information space given space in space the space question comma space the space required space circle space has space straight L subscript 1 space and space straight L subscript 2 space space as space its space tangents space and space also space touches space the space circle space 7 straight x squared minus 18 xy plus 7 straight y squared equals 0. Slopes space of space straight L subscript 1 space space and space space straight L subscript 2 space are space positive space as space straight m subscript 1 equals fraction numerator 1 over denominator 9 plus 4 square root of 2 end fraction space space and space straight m subscript 1 equals fraction numerator 1 over denominator 9 minus 4 square root of 2 end fraction So space both space space straight L subscript 1 space space and space space straight L subscript 2 space will space be space in space the space space first space quadrant. left square bracket Refer space figure right square bracket Let space straight O space be space the space centre space of space the space required space circle space left parenthesis straight S apostrophe right parenthesis AP minus Radius space of space straight S equals straight r space equals 4 square root of 2 OP space minus Radius space of space straight S apostrophe space equals space straight r apostrophe Since space straight S apostrophe space touches space straight S space internally space and space its space centred space away space from space left parenthesis 4 comma space 4 right parenthesis space as space in space fig. Now space space checking space options comma left parenthesis straight c right parenthesis space straight x squared plus straight y squared space equals space 8 space rightwards double arrow space centred space at space left parenthesis 0 comma space 0 right parenthesis Not space possible space space as space it space would space no space longer space have space space straight L subscript 1 space space and space space straight L subscript 2 space as space tangents left parenthesis straight b right parenthesis space straight x squared plus straight y squared space plus 12 straight x space plus space 12 straight y space plus 64 space equals space 0 space rightwards double arrow space centred space at space left parenthesis negative 6 comma space minus 6 right parenthesis Not space possible space space as space it space would space no space longer space be space contained space in space straight S. Option space left parenthesis straight a right parenthesis space space straight x squared plus straight y squared space minus 12 straight x space minus space 12 straight y space plus 64 space equals space 0 space rightwards double arrow space centred space at space left parenthesis 6 comma space 6 right parenthesis space space and space Radius space equals square root of 6 squared plus 6 squared minus 64 end root equals 2 square root of 2 space straight r minus straight r apostrophe space equals 4 square root of 2 minus 2 square root of 2 equals 2 square root of 2 space space left parenthesis positive space in space the space figure right parenthesis As space we space see space both space the space parameters space are space successfully space demonstrated space by space left parenthesis straight a right parenthesis So space left parenthesis straight a right parenthesis space must space be space the space answer. space
Answered by Vijaykumar Wani | 01 Dec, 2016, 03:21: PM

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