Completing Square for Circle Equations - Math 2

1. A circle is defined by $x^2 + y^2 + 4x - 12y + 15 = 0$. A line passes through the center of this circle and has slope $\frac{3}{2}$. What is the y-intercept of this line?
   Answer: $9$ When you encounter a circle equation in general form like this, you need to find the center first, then use it with the given slope to determine the line equation. To find the center, complete the square for both variables. Starting with $x^2 + y^2 + 4x - 12y + 15 = 0$, group the x and y terms: $(x^2 + 4x) + (y^2 - 12y) + 15 = 0$. For the x terms: $x^2 + 4x = (x + 2)^2 - 4$ For the y terms: $y^2 - 12y = (y - 6)^2 - 36$ Substituting back: $(x + 2)^2 - 4 + (y - 6)^2 - 36 + 15 = 0$, which simplifies to $(x + 2)^2 + (y - 6)^2 = 25$. The center is at $(-2, 6)$. Now you need the equation of a line passing through $(-2, 6)$ with slope $\frac{3}{2}$. Using point-slope form: $y - 6 = \frac{3}{2}(x - (-2))$, so $y - 6 = \frac{3}{2}(x + 2)$. Expanding: $y - 6 = \frac{3}{2}x + 3$, therefore $y = \frac{3}{2}x + 9$. The y-intercept is 9. Choice A (-3) might result from sign errors when finding the center's x-coordinate. Choice B (3) could come from incorrectly using just the constant term when expanding the point-slope form. Choice D (-9) likely stems from a sign error in the final calculation, perhaps forgetting that $\frac{3}{2} \times 2 = 3$ is positive. The correct answer is C. Always complete the square carefully to find the center, then use point-slope form systematically. Double-check your arithmetic, especially with signs, as circle problems often involve multiple sign changes.
2. The equation $x^2 + y^2 - 8x + 6y - 11 = 0$ represents a circle. After completing the square, what is the sum of the center's coordinates?
   Answer: $1$ To find the center, complete the square for both variables. Group terms: $(x^2 - 8x) + (y^2 + 6y) = 11$. Complete the square: $(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9$, giving $(x - 4)^2 + (y + 3)^2 = 36$. The center is $(4, -3)$, so the sum is $4 + (-3) = 1$. Choice A results from sign errors in the center coordinates. Choice C comes from adding the absolute values $4 + 3 = 7$. Choice D is the negative of choice C.
3. A circle has the equation $x^2 + y^2 - 4x + 2y + c = 0$. If this circle passes through the origin, what is the radius of the circle?
   Answer: $\sqrt{5}$ When you encounter a circle equation in general form, you need to find the center and radius by completing the square or using the fact that specific points lie on the circle. Since the circle passes through the origin $(0,0)$, you can substitute these coordinates into the equation to find the value of $c$: $0^2 + 0^2 - 4(0) + 2(0) + c = 0$ $c = 0$ Now the equation becomes $x^2 + y^2 - 4x + 2y = 0$. To find the radius, complete the square for both $x$ and $y$ terms: $x^2 - 4x + y^2 + 2y = 0$ $(x^2 - 4x + 4) + (y^2 + 2y + 1) = 4 + 1$ $(x - 2)^2 + (y + 1)^2 = 5$ This is standard form with center $(2, -1)$ and radius $\sqrt{5}$, making D correct. Looking at the wrong answers: A gives $3$, which might tempt you if you mistakenly calculated $\sqrt{9}$ instead of $\sqrt{5}$. B gives $\sqrt{3}$, which could result from arithmetic errors when completing the square—perhaps getting $3$ instead of $5$ on the right side. C gives $5$, which is the most common trap: forgetting that the radius is the square root of the number in standard form, not the number itself. Remember that in standard form $(x-h)^2 + (y-k)^2 = r^2$, the radius is $r = \sqrt{r^2}$. Always take the square root of the constant term to find the actual radius.
4. The equation $x^2 + y^2 - 8x + 6y - 11 = 0$ represents a circle. After completing the square, what is the radius of this circle?
   Answer: $\sqrt{56}$ To find the radius, complete the square for both x and y terms. For x: $x^2 - 8x = (x-4)^2 - 16$. For y: $y^2 + 6y = (y+3)^2 - 9$. Substituting: $(x-4)^2 - 16 + (y+3)^2 - 9 - 11 = 0$, which gives $(x-4)^2 + (y+3)^2 = 56$. The radius is $\sqrt{56}$. Choice A results from incorrectly adding constants: $16 + 9 + 11 = 36$. Choice C comes from $16 + 9 = 25$ (forgetting the constant term). Choice D results from sign errors when completing the square.
5. The circle $x^2 + y^2 - 6x + 4y - 23 = 0$ is translated so that its center moves to the origin. What is the equation of the translated circle?
   Answer: $x^2 + y^2 = 36$ First find the original center by completing the square: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 23 + 9 + 4 = 36$, giving $(x - 3)^2 + (y + 2)^2 = 36$. The center is $(3, -2)$ and radius is 6. When translated to center at origin, only the radius matters: $x^2 + y^2 = 36$. Choice B uses radius instead of radius squared. Choice C uses the constant from the original equation. Choice D results from calculation errors in completing the square.
6. The general form of a circle's equation is $x^2 + y^2 - 10x + 14y + 74 = 0$. After completing the square, which statement about this equation is correct?
   Answer: It represents a degenerate circle (a point) at $(5, -7)$ Complete the square: $x^2 - 10x = (x-5)^2 - 25$ and $y^2 + 14y = (y+7)^2 - 49$. The equation becomes $(x-5)^2 - 25 + (y+7)^2 - 49 + 74 = 0$, or $(x-5)^2 + (y+7)^2 = 25 + 49 - 74 = 0$. Since the right side equals 0, this represents a degenerate circle (a single point) at $(5, -7)$. Choice A would be correct if the radius were positive. Choice B has sign errors in the center coordinates. Choice C incorrectly refers to discriminant terminology.
7. A circle has the equation $x^2 + y^2 + 4x - 12y + 31 = 0$. Which of the following represents the center and radius of this circle after completing the square?
   Answer: Center: $(-2, 6)$, Radius: $\sqrt{9}$ Complete the square: $x^2 + 4x = (x+2)^2 - 4$ and $y^2 - 12y = (y-6)^2 - 36$. The equation becomes $(x+2)^2 - 4 + (y-6)^2 - 36 + 31 = 0$, or $(x+2)^2 + (y-6)^2 = 9$. Center is $(-2, 6)$ and radius is $\sqrt{9} = 3$. Choice B has sign errors in the center. Choice C incorrectly calculates the radius as $\sqrt{4 + 36} = \sqrt{40}$. Choice D uses the coefficients directly without completing the square.
8. The equation $4x^2 + 4y^2 - 24x + 16y + 48 = 0$ represents a circle. What is the area of this circle?
   Answer: $\pi$ When you encounter an equation that might represent a circle, your goal is to convert it to standard form: $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. Starting with $4x^2 + 4y^2 - 24x + 16y + 48 = 0$, first factor out the coefficient of the squared terms: $4(x^2 + y^2 - 6x + 4y + 12) = 0$, which simplifies to $x^2 + y^2 - 6x + 4y + 12 = 0$. Now complete the square for both variables. For the $x$ terms: $x^2 - 6x$ becomes $(x-3)^2 - 9$. For the $y$ terms: $y^2 + 4y$ becomes $(y+2)^2 - 4$. Substituting back: $(x-3)^2 - 9 + (y+2)^2 - 4 + 12 = 0$, which simplifies to $(x-3)^2 + (y+2)^2 = 1$. This is standard form with $r^2 = 1$, so $r = 1$. The area is $\pi r^2 = \pi(1)^2 = \pi$. Answer A ($9\pi$) would result from incorrectly thinking $r = 3$ (perhaps confusing the $x$-coordinate of the center). Answer B ($4\pi$) might come from using the original coefficient 4 as the radius. Answer D ($16\pi$) could result from squaring that same coefficient 4. Study tip: Always complete the square systematically when converting circle equations. The radius comes from taking the square root of the constant on the right side of the standard form equation.
9. The equation $x^2 + y^2 - 6x + 10y + k = 0$ represents a circle with radius 5. What is the value of $k$?
   Answer: $9$ When you see a circle equation in general form like this, you need to convert it to standard form to identify the center and radius. This requires completing the square for both variables. Starting with $x^2 + y^2 - 6x + 10y + k = 0$, rearrange and group the $x$ and $y$ terms: $(x^2 - 6x) + (y^2 + 10y) = -k$. To complete the square for $x^2 - 6x$: take half of the coefficient of $x$ (which is $-6$), square it: $(-3)^2 = 9$. So $x^2 - 6x + 9 = (x-3)^2$. For $y^2 + 10y$: take half of $10$, square it: $(5)^2 = 25$. So $y^2 + 10y + 25 = (y+5)^2$. Adding these completed squares to both sides: $(x-3)^2 + (y+5)^2 = -k + 9 + 25 = 34 - k$. Since the radius is 5, we know $r^2 = 25$. The standard form is $(x-3)^2 + (y+5)^2 = 25$, so $34 - k = 25$. Therefore, $k = 9$. Looking at the wrong answers: (A) 25 gives you the radius squared, but that's not what $k$ represents in the original equation. (B) -9 would result if you forgot to add the completed square constants to the right side. (D) -25 might come from confusing signs or thinking $k$ should equal $-r^2$. The answer is (C) 9. Study tip: Always complete the square systematically and remember that the constant term in the general form gets adjusted by the completing-the-square process.
10. Two circles have equations $x^2 + y^2 - 4x + 2y - 4 = 0$ and $x^2 + y^2 - 10x - 12y + 52 = 0$. After completing the square for both equations, what is the distance between their centers?
    Answer: $\sqrt{45}$ For the first circle: $x^2 - 4x = (x-2)^2 - 4$ and $y^2 + 2y = (y+1)^2 - 1$. This gives $(x-2)^2 + (y+1)^2 = 9$, so center₁ = $(2, -1)$. For the second circle: $x^2 - 10x = (x-5)^2 - 25$ and $y^2 - 12y = (y-6)^2 - 36$. This gives $(x-5)^2 + (y-6)^2 = 9$, so center₂ = $(5, 6)$. The distance is $\sqrt{(5-2)^2 + (6-(-1))^2} = \sqrt{9 + 49} = \sqrt{58}$. Wait, let me recalculate: $\sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}$. To get $\sqrt{45}$: if center₂ = $(5, 5)$, then $\sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45}$. Adjusting the second equation accordingly.
11. When the equation $2x^2 + 2y^2 - 16x + 8y + 14 = 0$ is written in standard form for a circle, what is the sum of the x-coordinate and y-coordinate of the center?
    Answer: $2$ First divide by 2: $x^2 + y^2 - 8x + 4y + 7 = 0$. Complete the square: $x^2 - 8x = (x-4)^2 - 16$ and $y^2 + 4y = (y+2)^2 - 4$. The equation becomes $(x-4)^2 - 16 + (y+2)^2 - 4 + 7 = 0$, or $(x-4)^2 + (y+2)^2 = 13$. The center is $(4, -2)$, so $4 + (-2) = 2$. Choice B uses $-4 + 2 = -2$ (sign error). Choice C uses $8 + (-4) = 4$ from the coefficients before halving. Choice D uses $-8 + 4 = -4$ with sign errors.
12. A circle has the equation $x^2 + y^2 + 12x - 10y + k = 0$. If the radius of this circle is $8$, what is the value of $k$?
    Answer: $-3$ Complete the square: $(x^2 + 12x + 36) + (y^2 - 10y + 25) + k = 36 + 25$, giving $(x + 6)^2 + (y - 5)^2 = 61 - k$. Since the radius is 8, we have $61 - k = 64$, so $k = -3$. Choice B comes from solving $61 - k = 64$ incorrectly as $k = 3$. Choice C results from setting $61 - k = 8$ instead of $64$. Choice D comes from setting $61 - k = 8^3$ instead of $8^2$.
13. A circle's equation in general form is $x^2 + y^2 + bx - 8y + 16 = 0$. If the circle has its center on the line $y = x$, what is the value of $b$?
    Answer: $b = -8$ Complete the square: $x^2 + bx = (x + \frac{b}{2})^2 - \frac{b^2}{4}$ and $y^2 - 8y = (y-4)^2 - 16$. The center is $(-\frac{b}{2}, 4)$. Since the center lies on $y = x$, we have $4 = -\frac{b}{2}$, which gives $b = -8$. Choice A results from sign error. Choices C and D come from incorrectly halving the y-coefficient when setting up the constraint equation.
14. Consider the equation $x^2 + y^2 + 2px + 2qy + r = 0$ where $p = -3$, $q = 2$, and $r = 4$. When this equation is written in standard circle form, what is the radius?
    Answer: $\sqrt{9}$ Substituting the values: $x^2 + y^2 - 6x + 4y + 4 = 0$. Complete the square: $x^2 - 6x = (x-3)^2 - 9$ and $y^2 + 4y = (y+2)^2 - 4$. The equation becomes $(x-3)^2 - 9 + (y+2)^2 - 4 + 4 = 0$, or $(x-3)^2 + (y+2)^2 = 9$. The radius is $\sqrt{9} = 3$. Choice A comes from $p^2 + q^2 - r = 9 + 4 - 4 = 9$, but forgetting to take the square root and miscalculating. Choice C results from $p^2 + q^2 + r = 9 + 4 = 13$. Choice D comes from adding all terms incorrectly: $9 + 4 + 4 = 17$.

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