Consider the following statements:Assertion ( | vedclass

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(A) For the circle $x^2 + y^2 = 1$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.

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Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

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(A) For the circle $x^2 + y^2 = 1$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.For a tangent to be parallel to the $x$-axis,the slope $\frac{dy}{dx}$ must be $0$.Setting $\frac{dy}{dx} = 0$ gives $x = 0$.Substituting $x = 0$ into the circle equation $x^2 + y^2 = 1$,we get $y^2 = 1$,so $y = \pm 1$.Thus,there are two points $(0, 1)$ and $(0, -1)$ where the tangent is parallel to the $x$-axis.Therefore,both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.

Consider the following statements:
Assertion $(A)$: The circle $x^2 + y^2 = 1$ has exactly two tangents parallel to the $x$-axis.
Reason $(R)$: $\frac{dy}{dx} = 0$ on the circle exactly at the points $(0, \pm 1)$.
Of these statements:

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Consider the following statements:Assertion $(A)$: The circle $x^2 + y^2 = 1$ has exactly two tangents parallel to the $x$-axis.Reason $(R)$: $\frac{dy}{dx} = 0$ on the circle exactly at the points $(0, \pm 1)$.Of these statements: