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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

Equation of the tangent to the circle, at the point \((1 , -1)\) whose centre is the point of intersection of the straight lines \(x - y = 1\) and \(2x + y= 3\) is

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

Answer & Solution

Correct Answer

(A) \(x + 4y+ 3 = 0\)

Step-by-step Solution

Detailed explanation

Point of intersection of lines \(x-y=1\) and \(2x+y=3\) is \(\left( {\frac{4}{3},\frac{1}{3}} \right)\) Slope of \(OP = \frac{{\frac{1}{3} + 1}}{{\frac{4}{3} - 1}} = \frac{{\frac{4}{3}}}{{\frac{1}{3}}} = 4\) Slope of tangent \( = - \frac{1}{4}\) Equation of tangent…