The area (in sq. units) of the triangle forme | vedclass

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(D) Given circle: $x^2+y^2-2gx-2hy+h^2=0$.Center $C = (g, h)$,radius $r = \sqrt{g^2+h^2-h^2} = |g|$.Length of tangent $OP = OQ = \sqrt{S_1} = \sqrt{0^

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$\frac{gh^3}{h^2+g^2}$ sq. units

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(D) Given circle: $x^2+y^2-2gx-2hy+h^2=0$.Center $C = (g, h)$,radius $r = \sqrt{g^2+h^2-h^2} = |g|$.Length of tangent $OP = OQ = \sqrt{S_1} = \sqrt{0^2+0^2-2g(0)-2h(0)+h^2} = |h|$.Area of $	riangle OPQ = \frac{1}{2} \cdot OP \cdot OQ \cdot \sin(2	heta) = \frac{1}{2} h^2 \sin(2	heta)$.In $	riangle OPC$,$\angle OPC = 90^{\circ}$,so $	an 	heta = \frac{PC}{OP} = \frac{|g|}{|h|}$.Then $\sin(2	heta) = \frac{2 	an 	heta}{1+	an^2 	heta} = \frac{2(|g|/|h|)}{1+(g^2/h^2)} = \frac{2|g||h|}{h^2+g^2}$.Area of $	riangle OPQ = \frac{1}{2} h^2 \cdot \frac{2|g||h|}{h^2+g^2} = \frac{|g|h^2|h|}{h^2+g^2} = \frac{|g|h^3}{h^2+g^2}$ sq. units.Assuming $g, h > 0$,the area is $\frac{gh^3}{h^2+g^2}$ sq. units.

The area (in sq. units) of the triangle formed by the two tangents drawn from the external point $O(0,0)$ to the circle $x^2+y^2-2gx-2hy+h^2=0$ and their chord of contact is

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