The equation of the circle having the lines y | vedclass

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Question

(A) The given equation of the normals is $y^2 - 2y + 4x - 2xy = 0$.This can be factored as $y(y - 2) - 2x(y - 2) = 0$,which gives $(y - 2)(y - 2x) = 0

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$x^2 + y^2 - 2x - 4y + 3 = 0$

Vedclass · Answer

(A) The given equation of the normals is $y^2 - 2y + 4x - 2xy = 0$.This can be factored as $y(y - 2) - 2x(y - 2) = 0$,which gives $(y - 2)(y - 2x) = 0$.Thus,the normals are the lines $y = 2$ and $y = 2x$.The intersection of these normals is the center $(h, k)$ of the circle.Solving $y = 2$ and $y = 2x$,we get $2 = 2x$,so $x = 1$.The center of the circle is $(1, 2)$.The general equation of a circle with center $(1, 2)$ is $(x - 1)^2 + (y - 2)^2 = r^2$,which simplifies to $x^2 + y^2 - 2x - 4y + 5 - r^2 = 0$.Since the circle passes through $(2, 1)$,we substitute these coordinates into the equation:$(2 - 1)^2 + (1 - 2)^2 = r^2$ $\Rightarrow 1^2 + (-1)^2 = r^2$ $\Rightarrow r^2 = 2$.Substituting $r^2 = 2$ into the circle equation: $x^2 + y^2 - 2x - 4y + 5 - 2 = 0$,which is $x^2 + y^2 - 2x - 4y + 3 = 0$.

The equation of the circle having the lines $y^2 - 2y + 4x - 2xy = 0$ as its normals and passing through the point $(2, 1)$ is:

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