Explanation:
D Let required point of intersection is \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) the equation of chord of contact with respect to given circle \(x^2+y^2-4 x-6 y+4=0\) is
\(\mathrm{xx}_1+\mathrm{yy}_1-2\left(\mathrm{x}+\mathrm{x}_1\right)-3\left(\mathrm{y}+\mathrm{y}_1\right)+4=0\)
\(\left(\mathrm{x}_1-2\right) \mathrm{x}+\left(\mathrm{y}_1-3\right) \mathrm{y}+\left(4-3 \mathrm{y}_1-2 \mathrm{x}_1\right)=0\)
Let the equation (i) represent the line, \(\mathrm{AB}, 4 \mathrm{x}+4 \mathrm{y}-11=0\) itself, then
\(\frac{\mathrm{x}_1-2}{4}=\frac{\mathrm{y}_1-3}{4}=\frac{4-3 \mathrm{y}_1-2 \mathrm{x}_1}{-11}=\mathrm{k}\)
Then,
\(\mathrm{x}_1=4 \mathrm{k}+2\)
\(\mathrm{y}_1=4 \mathrm{k}+3\)
\(2 \mathrm{x}_1+3 \mathrm{y}_1-4=11 \mathrm{k}\)
And
From equation (ii),
\(8 \mathrm{k}+4+12 \mathrm{k}+9-4=11 \mathrm{k}\)
\(9 \mathrm{k}+9=0 \Rightarrow \mathrm{k}=-1\)So, \(\quad\left(\mathrm{x}_1, \mathrm{y}_1\right)=(-2,-1)\)