match the columns
 

Asked by Sunil Soni | 10th Aug, 2016, 07:31: PM

Expert Answer:

begin mathsize 20px style For space the space straight D space subpart.
Consider space the space given space figure space shown space below. end style
 
 
 
begin mathsize 24px style We space can space see space that space OP space is space perpendicular space to space BC
Now space the space slope space of space OP space where space straight O left parenthesis 0 comma 0 right parenthesis space and space straight P left parenthesis straight a comma straight b right parenthesis space
equals space fraction numerator straight b minus 0 over denominator straight a minus 0 end fraction equals straight b over straight a
As space OP space perpendicular space BC space
rightwards double arrow slope space of space BC equals fraction numerator negative straight a over denominator straight b end fraction space space space space space space space open square brackets space if space straight m subscript 1 space is space slope space of space line space straight l subscript 1 space and space straight m subscript 2 space end subscript is space slope space of space line space straight l subscript 2 space then space straight m subscript 1 cross times straight m subscript 2 equals space 1 close square brackets
Now space equation space of space the space Chord space BC space with space straight P left parenthesis straight a comma space straight b right parenthesis space as space the space mid minus point space and space slope space fraction numerator negative straight a over denominator straight b end fraction space is
space rightwards double arrow straight y minus straight b equals space fraction numerator negative straight a over denominator straight b end fraction open parentheses straight x minus straight a close parentheses
space rightwards double arrow by plus ax equals straight a squared plus straight b squared space
Now space we space know space that space angle OPB space equals space 90 to the power of straight o space space and space straight P left parenthesis straight a comma space straight b right parenthesis space is space the space mid minus point space on space the space chord space BC.
rightwards double arrow OP thin space equals space rcos 45 to the power of straight o equals straight r cross times fraction numerator 1 over denominator square root of 2 end fraction equals fraction numerator straight r over denominator square root of 2 end fraction space space space space space space space space open square brackets straight r space is space the space radius space of space the space given space circle close square brackets
so comma space
straight a squared plus straight b squared space equals straight r squared over 2
Hence space the space locus space of space the space mid minus point space is space straight x squared plus straight y squared equals straight r squared over 2
Since space in space our space case space straight r equals space 2 space from space the space equation space of space space space straight x squared plus straight y squared equals 4
rightwards double arrow OP thin space equals space 2 cos 45 to the power of straight o equals 2 cross times fraction numerator 1 over denominator square root of 2 end fraction equals square root of 2
so comma space straight x squared plus straight y squared equals space open parentheses square root of 2 close parentheses squared equals 2 equals space straight K space space space space space

NOTE colon space Please space ask space one space query space at space straight a space time. end style

Answered by Tripti Agrawal | 11th Aug, 2016, 12:18: PM