Let x-y=0 and x+y=1 be two perpendicular diam | vedclass

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(B) The intersection of the two diameters $x-y=0$ and $x+y=1$ gives the center of the circle. Adding the two equations: $(x-y) + (x+y) = 0 + 1$ $\Righ

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

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(B) The intersection of the two diameters $x-y=0$ and $x+y=1$ gives the center of the circle. Adding the two equations: $(x-y) + (x+y) = 0 + 1$ $\Rightarrow 2x = 1$ $\Rightarrow x = \frac{1}{2}$. Substituting $x = \frac{1}{2}$ into $x-y=0$, we get $y = \frac{1}{2}$. So, the center of the circle is $(\frac{1}{2}, \frac{1}{2})$. Since the circle passes through the origin $(0, 0)$, the radius $R$ is the distance between the center $(\frac{1}{2}, \frac{1}{2})$ and the origin $(0, 0)$. $R = \sqrt{(\frac{1}{2}-0)^2 + (\frac{1}{2}-0)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Let $x-y=0$ and $x+y=1$ be two perpendicular diameters of a circle of radius $R$. The circle will pass through the origin if $R$ is equal to

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