The coordinates of the centre of the smallest | vedclass

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(D) For the smallest circle passing through the origin $(0, 0)$ with a diameter lying on the line $y = x + 1$,the centre of the circle must be the pro

Vedclass · Accepted Answer

$\left(-\frac{1}{2}, \frac{1}{2}\right)$

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(D) For the smallest circle passing through the origin $(0, 0)$ with a diameter lying on the line $y = x + 1$,the centre of the circle must be the projection of the origin onto the line $y = x + 1$.The line $y = x + 1$ can be written as $x - y + 1 = 0$.The line perpendicular to $x - y + 1 = 0$ passing through the origin $(0, 0)$ has the form $x + y = k$.Since it passes through $(0, 0)$,we have $0 + 0 = k$,so $k = 0$.The perpendicular line is $x + y = 0$,or $y = -x$.To find the centre,we find the intersection of $y = x + 1$ and $y = -x$.Substituting $y = -x$ into $y = x + 1$,we get $-x = x + 1$,which implies $2x = -1$,so $x = -\frac{1}{2}$.Then $y = -(-\frac{1}{2}) = \frac{1}{2}$.Thus,the centre of the circle is $\left(-\frac{1}{2}, \frac{1}{2}\right)$.

The coordinates of the centre of the smallest circle passing through the origin and having $y=x+1$ as a diameter are

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