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NCERT Solutions for Class 11 Mathematics Chapter 6 Linear Inequalities
  • NCERT Solution For Class 11 Mathematics

    Linear Inequalities Here is the CBSE Mathematics Chapter 6 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Mathematics Linear Inequalities Chapter 6 NCERT Solutions for Class 11 Mathematics Linear Inequalities Chapter 6 The following is a summary in Hindi and English for the academic year 2024-25. You can save these solutions to your computer or use the Class 11 Mathematics.

    Question 1
    CBSEENMA11015476

    Let C be the circle with centre at(1,1) and radius 1. If T is the circle centred at (0,y) passing through origin and touching the circle externally, then the radius of T is equal to

    • fraction numerator square root of 3 over denominator square root of 2 end fraction
    • fraction numerator square root of 3 over denominator 2 end fraction
    • 1/2

    • 1/4

    Solution

    D.

    1/4

    Let the coordinate of the centre of T be (0, K)
    Distance between their centre
    straight k plus 1 space equals space square root of 1 plus left parenthesis straight k minus 1 right parenthesis squared end root space left square bracket space because space straight C subscript 1 straight C subscript 2 space equals space straight k plus 1 right square bracket
rightwards double arrow straight k plus 1 space equals space square root of 1 plus straight k squared plus 1 minus 2 straight k end root
rightwards double arrow straight k plus 12 space equals space square root of straight k squared plus 2 minus 2 straight k end root
straight k squared space plus 1 plus 2 straight k space equals space straight k squared space plus 2 minus 2 straight k
rightwards double arrow space straight k space equals space 1 fourth
    So, the radius of circle T is k i.e, 1/4