A circle is drawn with the y-axis as a tangen | vedclass

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(C) The reflection of the point $(3, 4)$ in the line $y = x$ is obtained by swapping the coordinates,which gives the center of the circle as $(h, k) =

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$x^2 + y^2 - 8x - 6y + 9 = 0$

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(C) The reflection of the point $(3, 4)$ in the line $y = x$ is obtained by swapping the coordinates,which gives the center of the circle as $(h, k) = (4, 3)$.Since the circle is tangent to the $y$-axis,the radius $r$ of the circle is equal to the absolute value of the $x$-coordinate of the center,so $r = |h| = 4$.The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 4)^2 + (y - 3)^2 = 4^2$.Expanding this,we have $x^2 - 8x + 16 + y^2 - 6y + 9 = 16$.Simplifying,we get $x^2 + y^2 - 8x - 6y + 9 = 0$.

$A$ circle is drawn with the $y$-axis as a tangent and its center at the point which is the reflection of $(3, 4)$ in the line $y = x$. The equation of the circle is:

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