The equation of tangent to the circle $x^2+y^2=1$, which is perpendicular to the line $y=m x+1$; is

The equation of tangent to the circle $x^2+y^2=1$, which is perpendicular to the line $y=m x+1$; is
  1. $x+m y-\sqrt{1+m^2}=0$
  2. $m x+y-\sqrt{1+m^2}=0$
  3. $x+m y+\sqrt{1+m^2}=0$
  4. $m x+y+\sqrt{1+m^2}=0$

Solution

The equation of tangent to the circle $ x^2+y^2=1 \text { is } y=m_1 x+\sqrt{1+m_1^2} $ The line is perpendicular to the line $y=m x+1$ $ \begin{aligned} & y=\frac{-1}{m} x+\sqrt{1+\frac{1}{m^2}} \\ \Rightarrow & m y=-x+\sqrt{m^2+1} \\ \Rightarrow \quad & x+m y-\sqrt{m^2+1}=0 \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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