Maths-
General
Easy

Question

Polar of origin (0, 0) w.r.t. the circle x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0 touches the circle x squared plus y squared equals r squared, if

  1. straight c equals straight r open parentheses lambda squared plus mu squared close parentheses
  2. r equals c open parentheses lambda squared plus mu squared close parentheses
  3. c squared equals r squared open parentheses lambda squared plus mu squared close parentheses
  4. r squared equals c squared open parentheses lambda squared plus mu squared close parentheses

hintHint:

In this question, we have to find the correct option if Polar of origin (0, 0) w. r. t. the circle x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0 touches the circle x squared plus y squared equals r squared. As we know that the polar of equation is 0. So, using this and putting P (0,0), we can find a equation for perpendicular distance of a point P open parentheses x subscript 1 comma x subscript 2 close parentheses and center (0,0) we can find the value of r and solving that we will get the correct option.

The correct answer is: c squared equals r squared open parentheses lambda squared plus mu squared close parentheses


    E q u a t i o n space p o l o r space o f space t h e space o r i g i n space w. r. t. space t h e space c i r c l e space i s space
x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0
p o l a r space o f space e q u a t i o n space equals 0 comma space P space left parenthesis x subscript 1 comma y subscript 1 right parenthesis
N o w comma space x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0
rightwards double arrow x. x subscript 1 plus y. y subscript 1 plus 2. lambda open parentheses fraction numerator x plus x subscript 1 over denominator 2 end fraction close parentheses plus 2. mu open parentheses fraction numerator y plus y subscript 1 over denominator 2 end fraction close parentheses plus c equals 0
rightwards double arrow x. x subscript 1 plus y. y subscript 1 plus lambda left parenthesis x plus x subscript 1 right parenthesis plus mu left parenthesis y plus y subscript 1 right parenthesis plus c equals 0
rightwards double arrow x.0 plus y.0 plus lambda left parenthesis x plus 0 right parenthesis plus mu left parenthesis y plus 0 right parenthesis plus c equals 0
rightwards double arrow lambda x plus mu y plus c equals 0
G i v e n comma space x squared plus y squared equals r squared
rightwards double arrow left parenthesis x minus h right parenthesis squared plus left parenthesis y minus k right parenthesis squared equals r to the power of 2 end exponent p e r p e n d i c u l a r space d i s tan c e space b e t w e e n space a space p o i n t space left parenthesis x subscript 1 comma y subscript 1 right parenthesis a n d space a space l i n e space left parenthesis lambda x plus mu y plus c equals 0 right parenthesis
P equals fraction numerator open vertical bar lambda x subscript 1 plus mu y subscript 1 plus c close vertical bar over denominator square root of lambda squared plus mu squared end root end fraction rightwards double arrow r equals fraction numerator open vertical bar lambda.0 plus mu.0 plus c close vertical bar over denominator square root of lambda squared plus mu squared end root end fraction rightwards double arrow r equals fraction numerator c over denominator square root of lambda squared plus mu squared end root end fraction
s q u a r i n g space b o t h space s i d e s comma
rightwards double arrow r squared equals fraction numerator c squared over denominator lambda squared plus mu squared end fraction
rightwards double arrow r squared open parentheses lambda squared plus mu squared close parentheses equals c squared

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