If the tangent at a point P(x, y) of a curve | vedclass

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(A) Let the curve be $y = f(x)$. The slope of the line joining the origin $(0, 0)$ and the point $P(x, y)$ is $m_1 = \frac{y}{x}$.The slope of the tan

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Circle

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(A) Let the curve be $y = f(x)$. The slope of the line joining the origin $(0, 0)$ and the point $P(x, y)$ is $m_1 = \frac{y}{x}$.The slope of the tangent at $P(x, y)$ is $m_2 = \frac{dy}{dx}$.Since the tangent is perpendicular to the line joining the origin to $P$,we have $m_1 	imes m_2 = -1$.Therefore,$\frac{y}{x} 	imes \frac{dy}{dx} = -1$.This implies $y \, dy = -x \, dx$.Integrating both sides,we get $\int y \, dy = -\int x \, dx$,which gives $\frac{y^2}{2} = -\frac{x^2}{2} + C$.This simplifies to $x^2 + y^2 = 2C$,which represents a circle centered at the origin.

If the tangent at a point $P(x, y)$ of a curve is perpendicular to the line that joins the origin with the point $P$,then the curve is

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