Question:

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 $+ y = 3$is :

Updated On: June 02, 2025
  • $4x+y-3=0$
  • $x+4y+3=0$
  • $3x-y-4=0$
  • $x-3y-4=0$
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The Correct Option is B

Solution and Explanation

Paint of intersection of lines
$x - y = 1$ and $2x+y = 3$
o is o $\left(\frac{4}{3}, \frac{1}{3}\right)$
Slope of OP $= \frac{\frac{1}{3}+1}{\frac{4}{3}-1} = \frac{\frac{4}{3}}{\frac{1}{3}} = 4$
Slope of tangent $= -\frac{1}{4}$
slope of tangent $y +1 = -\frac{1}{4}\left(x-1\right)$
$4y+4=-x+4$
$x+4y+3 = 0$
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