The equation of the circle symmetric to the c | vedclass

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(A) The given circle is $x^2 + y^2 - 2x - 4y + 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -2, c = 4$. The center is $(

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$x^2 + y^2 - 10x + 4y + 28 = 0$

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(A) The given circle is $x^2 + y^2 - 2x - 4y + 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -2, c = 4$. The center is $(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 2^2 - 4} = \sqrt{1} = 1$. We need to find the image of the center $(1, 2)$ about the line $x - y - 3 = 0$. Let the image be $(a, b)$. Using the formula for the reflection of a point $(x_1, y_1)$ about the line $Ax + By + C = 0$: $\frac{a - x_1}{A} = \frac{b - y_1}{B} = -2 \frac{Ax_1 + By_1 + C}{A^2 + B^2}$ $\frac{a - 1}{1} = \frac{b - 2}{-1} = -2 \frac{1(1) - 1(2) - 3}{1^2 + (-1)^2}$ $\frac{a - 1}{1} = \frac{b - 2}{-1} = -2 \frac{-4}{2} = 4$ $a - 1 = 4 \Rightarrow a = 5$ $b - 2 = -4 \Rightarrow b = -2$ The new center is $(5, -2)$ and the radius remains $r = 1$. The equation of the circle is $(x - 5)^2 + (y + 2)^2 = 1^2$. $x^2 - 10x + 25 + y^2 + 4y + 4 = 1$ $x^2 + y^2 - 10x + 4y + 28 = 0$.

The equation of the circle symmetric to the circle $x^2 + y^2 - 2x - 4y + 4 = 0$ about the line $x - y = 3$ is

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