The circle passing through (1, -2) and touchi | vedclass

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Question

(C) Let the equation of the circle be $(x-3)^2 + (y-0)^2 + \lambda y = 0$.Since it passes through $(1, -2)$:$(1-3)^2 + (-2)^2 + \lambda(-2) = 0$$4 + 4

Vedclass · Accepted Answer

$(5, -2)$

Vedclass · Answer

(C) Let the equation of the circle be $(x-3)^2 + (y-0)^2 + \lambda y = 0$.Since it passes through $(1, -2)$:$(1-3)^2 + (-2)^2 + \lambda(-2) = 0$$4 + 4 - 2\lambda = 0$$8 = 2\lambda \Rightarrow \lambda = 4$.Thus,the equation of the circle is $(x-3)^2 + y^2 + 4y = 0$.Checking the options:For $(5, -2)$: $(5-3)^2 + (-2)^2 + 4(-2) = 2^2 + 4 - 8 = 4 + 4 - 8 = 0$.Therefore,the circle passes through $(5, -2)$.

The circle passing through $(1, -2)$ and touching the $x$-axis at $(3, 0)$ also passes through the point

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