AP EAMCET · Maths · Circle
The equation of tangent to the circle \(x^2+y^2=1\), which is perpendicular to the line \(y=m x+1\); is
- A \(x+m y-\sqrt{1+m^2}=0\)
- B \(m x+y-\sqrt{1+m^2}=0\)
- C \(x+m y+\sqrt{1+m^2}=0\)
- D \(m x+y+\sqrt{1+m^2}=0\)
Answer & Solution
Correct Answer
(A) \(x+m y-\sqrt{1+m^2}=0\)
Step-by-step Solution
Detailed explanation
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 \(y=m x+1\)…
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