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
- A \(x + 4y+ 3 = 0\)
- B \(3x - y- 4 = 0\)
- C \(x-3y-4 = 0\)
- D \(4x + y- 3 = 0\)
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…
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