If the line 4x + 4y - 11 = 0 intersects the c | vedclass

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(D) Let the required point of intersection be $(x_1, y_1)$. The equation of the chord of contact with respect to the circle $x^2 + y^2 - 4x - 6y + 4 =

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$(-2, -1)$

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(D) Let the required point of intersection be $(x_1, y_1)$. The equation of the chord of contact with respect to the circle $x^2 + y^2 - 4x - 6y + 4 = 0$ is given by $T = 0$:$xx_1 + yy_1 - 2(x + x_1) - 3(y + y_1) + 4 = 0$Rearranging the terms,we get:$(x_1 - 2)x + (y_1 - 3)y + (4 - 2x_1 - 3y_1) = 0$ $(i)$Since this represents the line $4x + 4y - 11 = 0$,the coefficients must be proportional:$\frac{x_1 - 2}{4} = \frac{y_1 - 3}{4} = \frac{4 - 2x_1 - 3y_1}{-11} = K$From this,we have $x_1 = 4K + 2$ and $y_1 = 4K + 3$.Substituting these into the third ratio:$\frac{4 - 2(4K + 2) - 3(4K + 3)}{-11} = K$$\frac{4 - 8K - 4 - 12K - 9}{-11} = K$$\frac{-20K - 9}{-11} = K$$-20K - 9 = -11K$$-9K = 9 \Rightarrow K = -1$Substituting $K = -1$ back into the expressions for $x_1$ and $y_1$:$x_1 = 4(-1) + 2 = -2$$y_1 = 4(-1) + 3 = -1$Thus,the point of intersection is $(-2, -1)$.

If the line $4x + 4y - 11 = 0$ intersects the circle $x^2 + y^2 - 4x - 6y + 4 = 0$ at $A$ and $B$,then the point of intersection of the tangents drawn at $A$ and $B$ is

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