The equation of the tangent to the curve give | vedclass

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(C) Given the parametric equations of the curve are $x=3 \cos 	heta$ and $y=3 \sin 	heta$. Squaring and adding these equations,we get $x^2+y^2 = 9(\

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$x+y=3\sqrt{2}$

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(C) Given the parametric equations of the curve are $x=3 \cos 	heta$ and $y=3 \sin 	heta$. Squaring and adding these equations,we get $x^2+y^2 = 9(\cos^2 	heta + \sin^2 	heta) = 9$,which represents a circle with radius $r=3$. The point of tangency at $	heta = \frac{\pi}{4}$ is given by $x_1 = 3 \cos(\frac{\pi}{4}) = \frac{3}{\sqrt{2}}$ and $y_1 = 3 \sin(\frac{\pi}{4}) = \frac{3}{\sqrt{2}}$. The equation of the tangent to the circle $x^2+y^2=r^2$ at point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. Substituting the values,we get $x(\frac{3}{\sqrt{2}}) + y(\frac{3}{\sqrt{2}}) = 9$. Multiplying both sides by $\frac{\sqrt{2}}{3}$,we obtain $x+y = 3\sqrt{2}$.

The equation of the tangent to the curve given by $x=3 \cos \theta, y=3 \sin \theta$ at $\theta=\frac{\pi}{4}$ is

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