Question

The straight line $x-y-3=0$ touches the circle $x^2+y^2-4x+6y+11=0$ at the point whose co-ordinates are?

• A. $(1,-2)$
• B. $(1,2)$
• C. $(-1,2)$
• D. $(-1,-2)$

Answer 1

The correct option is A $(1,-2)$

Let the line $x-y-3=0$ touches the given circle at point $(x_1,y_1)$

then,

Equation of the tangent at $(x_1,y_1)$ to the given circle is

$x{x}_1+y{y}_1-2(x+x_1)+3(y+y_1)+11=0$

$x_{1}x+y_{1}y-2xs-2x_1+3y+3y_1+11=0$

$(x_1-2)x+(y_1+3)y-2x_1+3y_1+11=0$

But the given tangent at $(x_1,y_1)$ is

$x-y-3=0$

therefore,

$\frac{x_1-2}{1}=\frac{y_1+3}{-1}=\frac{-2x_1+3y_1+11}{-3}$

$x_1=k+2$ --- $\left(1\right)$

$y_1=-k-3$ --- $\left(2\right)$

$-2x_1+3y_1+11=-3k$ --- $\left(3\right)$

Putting values of $x_1$ and $y_1$ from $1$ and $2$ in $3$.

$-2(k+2)+3(-k-3)+11=-3k$

$k=-1$

from $1$

$x_1=k+2=-1+2=1$

From $2$.

$y_1=-k-3=1-3=-2$

Hence, point of contact is $(1,-2)$

therefore,

option $A$ is correct answer.