2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main - Mock Test - 11

  • question_answer
    The equations of perpendicular bisectors of two sides AB and AC of a triangle ABC are \[x+y+1=0\] and \[x-y+1=0,\] respectively. If circumradius of \[\Delta ABC\] is 2 units, then locus of vertex A is

    A)  \[{{x}^{2}}+{{y}^{2}}+2x-3=0\]

    B) \[{{x}^{2}}+{{y}^{2}}+2x+3=0\]   

    C) \[{{x}^{2}}+{{y}^{2}}-2x+3=0\]    

    D) \[{{x}^{2}}+{{y}^{2}}-2x-3=0\]

    Correct Answer: A

    Solution :

    [a] Solving perpendicular bisector \[x-y+1=0\]and \[x+y+1=0,\]we get \[P(-1,0),\] which is circumcenter of the triangle ABC.  Let point A be \[(x,y)\] Circumradius = 2 units \[\therefore \text{ }AP=2\] or \[{{(x+1)}^{2}}+{{y}^{2}}={{2}^{2}}\] or \[{{x}^{2}}+{{y}^{2}}+2x-3=0\] 


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