The equation of the normal to the circle x^2 | vedclass

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(C) The equation of the normal to a circle $x^2 + y^2 = r^2$ at a point $(x_1, y_1)$ on the circle is given by the line passing through the center $(0

Vedclass · Accepted Answer

$x - y = 0$

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(C) The equation of the normal to a circle $x^2 + y^2 = r^2$ at a point $(x_1, y_1)$ on the circle is given by the line passing through the center $(0, 0)$ and the point $(x_1, y_1)$.The slope of the normal is $m = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1}$.Given the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$,the slope is $m = \frac{1/\sqrt{2}}{1/\sqrt{2}} = 1$.The equation of the line passing through $(0, 0)$ with slope $m = 1$ is $y - 0 = 1(x - 0)$,which simplifies to $y = x$ or $x - y = 0$.

The equation of the normal to the circle $x^2 + y^2 = 9$ at the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$ is:

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