JEE Main & Advanced Sample Paper JEE Main Sample Paper-23

  • question_answer
    A circle \[S=0\] passes through points of intersection of circles \[{{x}^{2}}+{{y}^{2}}-2x+4y=1\] and \[{{x}^{2}}+{{y}^{2}}-4x-2y-5=0\] and cuts the circle \[{{x}^{2}}+{{y}^{2}}-4=0\] orthogonally. Then the length of tangent from origin on circle \[S=0\], is

    A)  3                                

    B)  2

    C)  1                                

    D)  4

    Correct Answer: B

    Solution :

    Equation of required circle is \[{{x}^{2}}+{{y}^{2}}+\frac{2(2\lambda -1)}{\lambda +1}\,x+\frac{2(2-\lambda )}{1+\lambda }\,y-\frac{(1+5\lambda )}{1+\lambda }=0\] Now, using condition of orthogonality, we get \[-4=\frac{1+5\lambda }{1+\lambda }\,\Rightarrow \,\lambda =\frac{-5}{9}\] So, the equation of circle is \[{{x}^{2}}+{{y}^{2}}\,-\frac{19}{2}\,x+\,\frac{23}{2}\,y+4=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner