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AP EAMCET202017 Sep 2020Evening ShiftMathematicsCircleActual

The equation of normal at ((1,1) ) to the circle (x^2+y^2-x-3 y-4=0 ) is

Options

  1. A(x+y-2=0 )
  2. B(2 x-y-1=0 )
  3. C(x-y+2=0 )
  4. D(x-y-2=0 )

Correct answer

A. (x+y-2=0 )

Step-by-step solution

Circle is, (x^2+y^2-x-3 y-4=0 ) Slope of tangent at ((1,1) ) is obtained by differentiating above equation, ( aligned & 2 x+2 y y^ -1-3 y^ =0 & y^ = . 1-2 x 2 y-3 |_ (1,1) & y^ = 1-2 2-3 =1 aligned ) So, slope of normal is, (m_N=-1 / m_T=-1 ) Equation of normal in one point forms is, ( aligned & y-y₁=m (x-x₁ ) & y-1=-1(x-1) & x+y-2=0 aligned )

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