Circle Equations - Math 3

1. A circle has equation $x^2 + y^2 - 8x + 2y + 8 = 0$. If this circle is reflected across the line $y = x$, what is the equation of the resulting circle?
   Answer: $x^2 + y^2 + 2x - 8y + 8 = 0$ First find the center of the original circle by completing the square: $(x-4)^2 + (y+1)^2 = 16 + 1 - 8 = 9$. The center is $(4, -1)$ with radius $3$. When reflected across $y = x$, the point $(a, b)$ becomes $(b, a)$. So the new center is $(-1, 4)$. The radius remains $3$. The new equation is $(x+1)^2 + (y-4)^2 = 9$, which expands to $x^2 + 2x + 1 + y^2 - 8y + 16 = 9$, giving $x^2 + y^2 + 2x - 8y + 8 = 0$. Choice B has wrong signs. Choice C doesn't reflect properly. Choice D changes the wrong coefficients.
2. Two concentric circles have equations $x^2 + y^2 - 6x + 8y = 0$ and $x^2 + y^2 - 6x + 8y + k = 0$. If the area of the annulus (ring) between them is $21\pi$, what is the value of $k$?
   Answer: $-21$ Complete the square for the first circle: $(x-3)^2 + (y+4)^2 = 9 + 16 = 25$, so $r_1 = 5$. For the second circle: $(x-3)^2 + (y+4)^2 = 25 - k$, so $r_2 = \sqrt{25-k}$. The annulus area is $\pi r_1^2 - \pi r_2^2 = \pi(25) - \pi(25-k) = \pi k$. Setting this equal to $21\pi$ gives $k = 21$. But this assumes $r_1 > r_2$, which requires $k > 0$. If $k < 0$, then $r_2 > r_1$ and the area is $\pi(25-k) - 25\pi = -\pi k = 21\pi$, so $k = -21$. Choice A is correct.
3. A circle intersects the coordinate axes at exactly three points. If two of these points are $(-4, 0)$ and $(0, 3)$, and the circle passes through the origin, what is the $x$-coordinate of the center?
   Answer: $-2$ Since the circle passes through $(0,0)$, $(-4,0)$, and $(0,3)$, we can use the general form $x^2 + y^2 + Dx + Ey + F = 0$. From $(0,0)$: $F = 0$. From $(-4,0)$: $16 - 4D = 0$, so $D = 4$. From $(0,3)$: $9 + 3E = 0$, so $E = -3$. The equation is $x^2 + y^2 + 4x - 3y = 0$. Completing the square: $(x+2)^2 + (y-\frac{3}{2})^2 = 4 + \frac{9}{4} = \frac{25}{4}$. The center is $(-2, \frac{3}{2})$, so the $x$-coordinate is $-2$. Choice A uses wrong arithmetic. Choice C confuses $x$ and $y$ coordinates. Choice D is off by 1.
4. A circle passes through points $A(0, 0)$, $B(6, 0)$, and $C(3, 4)$. What is the radius of this circle?
   Answer: $\frac{25}{8}$ Let the circle have equation $(x-h)^2 + (y-k)^2 = r^2$. Using the three points: From A(0,0): $h^2 + k^2 = r^2$. From B(6,0): $(6-h)^2 + k^2 = r^2$. From C(3,4): $(3-h)^2 + (4-k)^2 = r^2$. Setting the first two equal: $h^2 + k^2 = (6-h)^2 + k^2$, so $h^2 = 36 - 12h + h^2$, giving $12h = 36$ and $h = 3$. Setting the first and third equal: $h^2 + k^2 = (3-h)^2 + (4-k)^2$. Substituting $h = 3$: $9 + k^2 = 0 + (4-k)^2 = 16 - 8k + k^2$, so $9 = 16 - 8k$, giving $8k = 7$ and $k = \frac{7}{8}$. The radius is $r = \sqrt{h^2 + k^2} = \sqrt{9 + \frac{49}{64}} = \sqrt{\frac{576 + 49}{64}} = \sqrt{\frac{625}{64}} = \frac{25}{8}$.

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