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(C) Given the curve equation $x^2 + y^2 = a^2$.Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$.Thus,the slope of the tangent is

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

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(C) Given the curve equation $x^2 + y^2 = a^2$.Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$.Thus,the slope of the tangent is $\frac{dy}{dx} = -\frac{x}{y}$.At the point $\left( \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right)$,the slope is $\frac{dy}{dx} = -\frac{a/\sqrt{2}}{a/\sqrt{2}} = -1$.The equation of the tangent is given by $y - y_1 = m(x - x_1)$.Substituting the values: $y - \frac{a}{\sqrt{2}} = -1 \left( x - \frac{a}{\sqrt{2}} \right)$.$y - \frac{a}{\sqrt{2}} = -x + \frac{a}{\sqrt{2}}$.$x + y = \frac{2a}{\sqrt{2}}$.$x + y = a\sqrt{2}$.

What is the equation of the tangent to the curve $x^2 + y^2 = a^2$ at the point $\left( \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right)$?

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