The number of circles passing through the ori | vedclass

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(B) Let the center of the circle be $(h, k)$. Since the circle passes through $(0,0)$, its radius $r$ is $\sqrt{h^2 + k^2}$.The circle touches the lin

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$2$

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(B) Let the center of the circle be $(h, k)$. Since the circle passes through $(0,0)$, its radius $r$ is $\sqrt{h^2 + k^2}$.The circle touches the lines $x + y - 1 = 0$ and $x - y - 1 = 0$. Thus, the perpendicular distance from $(h, k)$ to these lines must equal $r$.$r = \frac{|h + k - 1|}{\sqrt{1^2 + 1^2}} = \frac{|h - k - 1|}{\sqrt{1^2 + (-1)^2}}$.This implies $|h + k - 1| = |h - k - 1|$.Case $1$: $h + k - 1 = h - k - 1 \implies 2k = 0 \implies k = 0$.Substituting $k=0$ into $r^2 = h^2 + k^2$, we get $r^2 = h^2$. Also $r = \frac{|h - 1|}{\sqrt{2}}$, so $h^2 = \frac{(h-1)^2}{2} \implies 2h^2 = h^2 - 2h + 1 \implies h^2 + 2h - 1 = 0$. This gives two values for $h$, so two circles.Case $2$: $h + k - 1 = -(h - k - 1) \implies h + k - 1 = -h + k + 1 \implies 2h = 2 \implies h = 1$.Substituting $h=1$ into $r^2 = h^2 + k^2$, we get $r^2 = 1 + k^2$. Also $r = \frac{|1 + k - 1|}{\sqrt{2}} = \frac{|k|}{\sqrt{2}}$, so $r^2 = \frac{k^2}{2}$.Equating these, $1 + k^2 = \frac{k^2}{2} \implies \frac{k^2}{2} = -1$, which has no real solution for $k$.Thus, there are $2$ such circles.

The number of circles passing through the origin $(0,0)$ and touching the lines $x + y = 1$ and $x - y = 1$ is ...

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