Question

The equation of the tangent from the point $\left(0,1\right)$ to the circle $x^2+y^2-2x-6y+6=0$, is

• A. $3\left(x^2-y^2\right)+4xy-4x-6y+3=0$
• B. $3y^2+4xy-4x-6y+3=0$
• C. $3x^2+4xy-4x-6y+3=0$
• D. $3\left(x^2+y^2\right)+4xy-4x-6y+3=0$

Answer 1

The correct option is B

$3y^2+4xy-4x-6y+3=0$

Explanation for the correct option

Given circle is $x^2+y^2-2x-6y+6=0$ and given point is $\left(0,1\right)$
So, $x_1=0,y_1=1$

The equation of pairs of tangents to the circle $x^2+y^2+2gx+2fy+c=0$ from a point $\left(x_1,y_1\right)$ is given as,
$\left(x^2+y^2+2gx+2fy+c\right)\left({x_1}^2+{y_1}^2+2g{x}_1+2f{y}_1+c\right)={\left[x·x_1+y·y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c\right]}^2\dots \left(1\right)$

Comparing the given equation to the $x^2+y^2+2gx+2fy+c=0$ form of circle, we see that,
$\begin{array}{cc}& 2gx=-2x\\ \Rightarrow & g=-1\\ & 2fy=-6y\\ \Rightarrow & f=-3\\ & c=6\end{array}$

Substituting the values in $\left(1\right)$,
$\begin{array}{cc}\Rightarrow & \left[x^2+y^2+2\left(-1\right)x+2\left(-3\right)y+6\right]\left[0^2+1^2+2\left(-1\right)\left(0\right)+2\left(-3\right)\left(1\right)+6\right]={\left[0·x+1·y+\left(-1\right)\left(x+0\right)+\left(-3\right)\left(y+1\right)+6\right]}^2\\ \Rightarrow & \left(x^2+y^2-2x-6y+6\right)\left(0+1+0-6+6\right)={\left(y-x-3y-3+6\right)}^2\\ \Rightarrow & \left(x^2+y^2-2x-6y+6\right)\left(1\right)={\left(-x-2y+3\right)}^2\\ \Rightarrow & x^2+y^2-2x-6y+6=x^2+4y^2+9+4xy-6x-12y\left[\because {\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ca\right]\\ \Rightarrow & \cancel{x^2}+y^2-2x-6y+6=\cancel{x^2}+4y^2+4xy-6x-12y+9\\ \Rightarrow & 3y^2+4xy-4x-6y+3=0\end{array}$

Therefore, the equation of the pair of tangents is $3y^2+4xy-4x-6y+3=0$

Hence, option(B) i.e. $3y^2+4xy-4x-6y+3=0$ is correct.