The locus of the centre of circles which pass | vedclass

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Question

(A) Let the centre of the circle be $(h, k)$.Since the circle passes through $(0, 1)$,the radius $r$ is the distance between $(h, k)$ and $(0, 1)$,so

Vedclass · Accepted Answer

$(x + y)^2 = 4y - 2$

Vedclass · Answer

(A) Let the centre of the circle be $(h, k)$.Since the circle passes through $(0, 1)$,the radius $r$ is the distance between $(h, k)$ and $(0, 1)$,so $r^2 = h^2 + (k - 1)^2$.Since the circle touches the line $y - x = 0$,the radius $r$ is the perpendicular distance from $(h, k)$ to the line $y - x = 0$,so $r = \frac{|k - h|}{\sqrt{1^2 + (-1)^2}} = \frac{|k - h|}{\sqrt{2}}$.Equating the two expressions for $r^2$:$h^2 + (k - 1)^2 = \frac{(k - h)^2}{2}$$2(h^2 + k^2 - 2k + 1) = k^2 + h^2 - 2hk$$2h^2 + 2k^2 - 4k + 2 = k^2 + h^2 - 2hk$$h^2 + k^2 + 2hk = 4k - 2$$(h + k)^2 = 4k - 2$Replacing $(h, k)$ with $(x, y)$,the locus is $(x + y)^2 = 4y - 2$.

The locus of the centre of circles which pass through $(0, 1)$ and touch the line $y = x$ is -

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