If PA and PB are the tangents drawn from the | vedclass

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(C) Given,from point $P(1,1)$,tangents $PA$ and $PB$ are drawn to the circle $x^2+y^2+gx+gy-2=0$.The length of the tangent $PA = \sqrt{S_1} = \sqrt{1^

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$\sqrt{g^3+4g}$

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(C) Given,from point $P(1,1)$,tangents $PA$ and $PB$ are drawn to the circle $x^2+y^2+gx+gy-2=0$.The length of the tangent $PA = \sqrt{S_1} = \sqrt{1^2+1^2+g(1)+g(1)-2} = \sqrt{1+1+g+g-2} = \sqrt{2g}$.The centre $C$ of the circle is $(-\frac{g}{2}, -\frac{g}{2})$ and the radius $r = AC = \sqrt{(-\frac{g}{2})^2 + (-\frac{g}{2})^2 - (-2)} = \sqrt{\frac{g^2}{4} + \frac{g^2}{4} + 2} = \sqrt{\frac{g^2}{2} + 2} = \sqrt{\frac{g^2+4}{2}} = \frac{\sqrt{g^2+4}}{\sqrt{2}}$.Since $PA$ is a tangent,$\angle PAC = 90^\circ$. The area of $	riangle PAC = \frac{1}{2} 	imes PA 	imes AC = \frac{1}{2} 	imes \sqrt{2g} 	imes \frac{\sqrt{g^2+4}}{\sqrt{2}} = \frac{1}{2} \sqrt{g(g^2+4)} = \frac{1}{2} \sqrt{g^3+4g}$.The quadrilateral $PACB$ consists of two congruent triangles $	riangle PAC$ and $	riangle PBC$.Therefore,the area of quadrilateral $PACB = 2 	imes 	ext{Area}(	riangle PAC) = 2 	imes \frac{1}{2} \sqrt{g^3+4g} = \sqrt{g^3+4g}$.

If $PA$ and $PB$ are the tangents drawn from the point $P(1,1)$ to the circle $x^2+y^2+gx+gy-2=0$ with $C$ as the centre,then the area (in sq. units) of the quadrilateral $PACB$ is

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