The equation of the normal to a circle at the | vedclass

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(C) The normal to a circle at any point always passes through the center of the circle. Given the normal equation $y - x + 3 = 0$,we can rewrite it as

Vedclass · Accepted Answer

$(x - 3)^2 + y^2 = 9$

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(C) The normal to a circle at any point always passes through the center of the circle. Given the normal equation $y - x + 3 = 0$,we can rewrite it as $y = x - 3$. For a circle $(x - h)^2 + (y - k)^2 = r^2$,the center is $(h, k)$. Testing the centers of the given options: For option $(c)$,the circle is $(x - 3)^2 + y^2 = 9$,which has the center $(3, 0)$. Substituting the center $(3, 0)$ into the normal equation $y - x + 3 = 0$: $0 - 3 + 3 = 0$. Since the center satisfies the equation,the normal passes through the center. Thus,option $(c)$ is the correct circle.

The equation of the normal to a circle at the point $\left( 3 + \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right)$ is $y - x + 3 = 0$. To which of the following circles does this normal belong?

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