Tangents are drawn from (4, 4) to the circle | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 2x - 2y - 7 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (1, 1)$ an

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$3\sqrt{2}$

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(B) The equation of the circle is $x^2 + y^2 - 2x - 2y - 7 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (1, 1)$ and radius $R = \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 1^2 - (-7)} = \sqrt{9} = 3$.The equation of the chord of contact $AB$ from point $(x_1, y_1) = (4, 4)$ is given by $T = 0$,which is $xx_1 + yy_1 - (x + x_1) - (y + y_1) - 7 = 0$.Substituting the values,we get $4x + 4y - (x + 4) - (y + 4) - 7 = 0$,which simplifies to $3x + 3y - 15 = 0$ or $x + y - 5 = 0$.The distance $d$ from the center $(1, 1)$ to the chord $x + y - 5 = 0$ is $d = \frac{|1 + 1 - 5|}{\sqrt{1^2 + 1^2}} = \frac{|-3|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.The length of the chord $AB$ is $2\sqrt{R^2 - d^2} = 2\sqrt{3^2 - (\frac{3}{\sqrt{2}})^2} = 2\sqrt{9 - \frac{9}{2}} = 2\sqrt{\frac{9}{2}} = 2 	imes \frac{3}{\sqrt{2}} = 3\sqrt{2}$.

Tangents are drawn from $(4, 4)$ to the circle $x^2 + y^2 - 2x - 2y - 7 = 0$ to meet the circle at $A$ and $B$. The length of the chord $AB$ is

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