A tangent is drawn to the circle 2x^{2} + 2y^ | vedclass

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(B) The equation of the circle is $2x^{2} + 2y^{2} - 3x + 4y = 0$. Dividing by $2$,we get the standard form: $x^{2} + y^{2} - \frac{3}{2}x + 2y = 0$.

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$2$

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(B) The equation of the circle is $2x^{2} + 2y^{2} - 3x + 4y = 0$. Dividing by $2$,we get the standard form: $x^{2} + y^{2} - \frac{3}{2}x + 2y = 0$. Here,$AB$ represents the length of the tangent from point $B(2, 1)$ to the circle. The length of the tangent from a point $(x_{1}, y_{1})$ to a circle $x^{2} + y^{2} + 2gx + 2fy + c = 0$ is given by $\sqrt{x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c}$. Substituting the coordinates of $B(2, 1)$ into the circle equation: $AB = \sqrt{(2)^{2} + (1)^{2} - \frac{3}{2}(2) + 2(1)}$ $AB = \sqrt{4 + 1 - 3 + 2}$ $AB = \sqrt{4} = 2 	ext{ units}$.

$A$ tangent is drawn to the circle $2x^{2} + 2y^{2} - 3x + 4y = 0$ at point $A$ and it meets the line $x + y = 3$ at $B(2, 1)$. Then,the length of $AB$ is equal to:

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