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 $2x+y=3$ is :

• A. $x+4y+3=0$
• B. $x-3y-4=0$
• C. $4x+y-3=0$
• D. $3x-y-4=0$

Answer 1

The correct option is A $x+4y+3=0$

Solving equation $x-y=1\&2x+y=3$, we get, $x=\frac{4}{3},y=\frac{1}{3}$
$\Rightarrow$ Centre of the circle$=(\frac{4}{3},\frac{1}{3})$
Slope of radius $\left(OP\right)=\frac{-1-\frac{1}{3}}{1-\frac{4}{3}}=4$
Since, tangent is perpendicular to radius.
$\therefore$ Slope of tangent=$-\frac{1}{4}$
$\Rightarrow$ Equation of the tangent at $(1,-1)$ is,
$y+1=-\frac{1}{4}(x-1)\Rightarrow x+4y+3=0$

OR,
Equation of the circle is,
$(x-\frac{4}{3}{)}^2+(y-\frac{1}{3}{)}^2=(1-\frac{4}{3}{)}^2+(-1-\frac{1}{3}{)}^2\Rightarrow 3x^2+3y^2-8x-2y=0$
Hence, equation of the tangent at $(1,-1)$ is,
$3x-3y-4(x+1)-(y-1)=0\Rightarrow x+4y+3=0$