Circles & Parabolas - Algebra 3

1. For which value(s) of $k$ does $x^2+y^2+6x-4y+k=0$ represent a circle?
   Answer: $k<13$ Whenever you see a general second-degree equation like $x^2+y^2+6x-4y+k=0$, your first move should be to complete the square to reveal the circle's center and radius. Group the $x$ terms and $y$ terms: $(x^2+6x)+(y^2-4y)+k=0$. Completing the square gives $(x+3)^2-9+(y-2)^2-4+k=0$, so $(x+3)^2+(y-2)^2=13-k$. This is a circle exactly when the right side is positive, because the radius squared must be greater than zero. Thus you need $13-k>0$, which simplifies to $k<13$. Now look at the other choices. The choice $k>13$ would make $13-k$ negative, so no real points satisfy the equation — that gives an empty graph, not a circle. The choice $k\le 13$ includes too much: when $k=13$, the equation becomes $(x+3)^2+(y-2)^2=0$, which is only the single point $(-3,2)$, not a circle. The choice $k=13$ is exactly that degenerate point case, so it fails the requirement of having a positive radius. Your takeaway: after completing the square, always check the radius-squared constant. For a circle, it must be strictly positive. If it equals zero, you get a point; if it is negative, there is no graph. On this exam, the trap is forgetting that $k=13$ produces a degenerate case, so remember the strict inequality.
2. A parabola has vertex $(3,-1)$ and focus $(3,-4)$. Which statement is correct?
   Answer: Its equation is $(x-3)^2=-12(y+1)$ and its directrix is $y=2$. The focus is below the vertex, so the parabola opens downward with $p=-3$. Thus $(x-3)^2=4p(y+1)=-12(y+1)$, and the directrix is $y=k-p=-1-(-3)=2$. Choice B has the wrong directrix sign, C uses a horizontal-axis equation, and D uses the wrong sign for $p$.
3. A student claims that the circle $x^2+y^2+6x-10y+30=0$ has center $(-3,5)$ and radius $30$. Which statement correctly corrects the student?
   Answer: The center is $(-3,5)$, but the radius is $2$. Completing the square gives $(x+3)^2+(y-5)^2=4$, so the center is indeed $(-3,5)$, but the radius is $\sqrt4=2$, not $30$. Choice A has the wrong center signs, B treats the original constant as $r^2$, and D is false because $r^2=4$.
4. Which statement correctly describes the graph of $x^2+y^2+2x-4y+6=0$?
   Answer: It has no real graph because $(x+1)^2+(y-2)^2=-1$. Completing the square gives $(x+1)^2+(y-2)^2=-1$. Since the left side is a sum of squares and the right side is negative, no real ordered pairs satisfy the equation, so the graph is empty. The center $(-1,2)$ appears in several distractors, but the radius cannot be $1$ or $\sqrt6$, and the point $(-1,2)$ would be the graph only if the completed-square constant were zero.
5. Which equation represents the parabola with focus $(2,-1)$ and directrix $y=3$?
   Answer: $(x-2)^2=-8(y-1)$ The vertex is halfway between the focus and directrix: $(2,1)$. Since the focus is below the vertex, the directed distance is $p=-2$, so $4p=-8$. Thus the equation is $(x-2)^2=-8(y-1)$. Choice A uses the focus as the vertex, C uses a horizontal-axis equation, and D has the wrong sign for $p$.
6. Which equation represents the circle that passes through $(2,0)$, $(0,4)$, and $(-2,0)$?
   Answer: $x^2+\left(y-\frac32\right)^2=\frac{25}{4}$ By symmetry, the center is on the $y$-axis, so write it as $(0,k)$. Setting distances to $(2,0)$and $(0,4)$ equal gives $4+k^2=(4-k)^2$, so $k=3/2$. The radius squared is $4+(3/2)^2=25/4$. The center $(0,5/2)$ and the center $(3/2,0)$ do not pass through all three points, and the equation with radius squared $5/2$ uses the radius instead of the squared radius.
7. A parabola is given by $x=\frac{1}{2}(y-1)^2+3$. What are its vertex and focus?
   Answer: Vertex $(3,1)$; focus $(\frac{7}{2},1)$ Whenever you see a parabola written as $x=a(y-k)^2+h$, the roles of $x$ and $y$ are swapped compared with $y=a(x-h)^2+k$. The vertex is $(h,k)$, and the parabola opens right if $a>0$, left if $a<0$. Here $h=3$ and $k=1$, so the vertex is $(3,1)$. Rewrite the equation to find the focal length: $$x=\frac12(y-1)^2+3 \implies (y-1)^2=2(x-3).$$ This matches $(y-k)^2=4p(x-h)$, so $4p=2$, giving $p=\frac12$. Since the parabola opens right, the focus is $\frac12$ unit to the right of the vertex: $$(3+\tfrac12,1)=\left(\frac72,1\right).$$ So the vertex $(3,1)$; focus $(\frac72,1)$ is the correct pairing. The focus $(3,\frac32)$ treats the parabola as if it opened upward, but since $y$ is the squared variable, the focus must move horizontally, not vertically. The focus $(\frac52,1)$ is actually the directrix line, which lies $\frac12$ unit to the left of the vertex. The vertex $(-3,-1)$ uses the opposite signs for both the horizontal and vertical shifts; $+3$ and $-1$ in the equation mean $(3,1)$, not $(-3,-1)$. Its focus also has the wrong $x$-coordinate. On the exam, when a parabola has $x$ isolated and $y$ squared, think "sideways": vertex and focus share the same $y$-coordinate, and the focus is $p$ units right or left of the vertex.
8. The equation $2x^2+2y^2-12x+8y+10=0$ represents a circle. Which statement correctly identifies its center and radius?
   Answer: Center $(3,-2)$, radius $2\sqrt2$ Divide the equation by 2 to get $x^2+y^2-6x+4y+5=0$. Completing the square gives $(x-3)^2+(y+2)^2=8$, so the center is $(3,-2)$ and the radius is $\sqrt8=2\sqrt2$. The other choices either use the wrong center signs or treat the constant as the radius squared.
9. A circle is centered at $(1,-3)$ and is tangent to the line $4x-3y=7$. Which equation represents this circle?
   Answer: $(x-1)^2+(y+3)^2=\frac{36}{25}$ When you see a circle tangent to a line, the key fact is that the radius equals the perpendicular distance from the center to that tangent line. So here, start with the center $(1,-3)$, which gives the circle equation form $(x-1)^2+(y+3)^2=r^2$. To find $r$, use the distance from a point to a line: $\frac{|4(1)-3(-3)-7|}{\sqrt{4^2+(-3)^2}}=\frac{|4+9-7|}{5}=\frac{6}{5}$. That distance is the radius, so the right side must be $r^2$, not $r$: $\left(\frac{6}{5}\right)^2=\frac{36}{25}$. Thus the correct equation is $(x-1)^2+(y+3)^2=\frac{36}{25}$. Now look at the traps in the other choices. The equation with $(x-1)^2+(y+3)^2=36$ uses the correct center but treats $6/5$ as if it were $6$, forgetting to divide by the square-root denominator in the distance formula. The equation with (x-1)^2+(y+3)^2=\frac{6}{5}} uses the correct center but puts the radius itself on the right instead of the radius squared; the right side of a circle equation is always $r^2$. The equation with $(x+1)^2+(y-3)^2=\frac{36}{25}$ has the correct radius squared but the wrong center signs — it corresponds to center $(-1,3)$, not $(1,-3)$. Your study takeaway: whenever a circle is tangent to a line, immediately set radius = distance from center to line, then square that distance for the equation. Also double-check that the signs inside $(x-h)$ and $(y-k)$ match the center coordinates.
10. The circle $(x-2)^2+(y+3)^2=25$ has a diameter with one endpoint at $(5,1)$. What are the coordinates of the other endpoint of the diameter?
    Answer: $(-1,-7)$ The center $(2,-3)$ is the midpoint of the diameter. If the other endpoint is $(x,y)$, then $(5+x)/2=2$ and $(1+y)/2=-3$, giving $x=-1$ and $y=-7$. Choice A flips the sign of the y-coordinate, C flips the sign of the x-coordinate, and D adds the center and endpoint instead of using the midpoint relation.
11. A circle has a diameter whose endpoints are $(1,-2)$ and $(7,4)$. Which equation represents this circle?
    Answer: $(x-4)^2+(y-1)^2=18$ Whenever you see a circle defined by a diameter, your first move is to find the center and the radius. The center is the midpoint of the endpoints: $\left(\frac{1+7}{2},\frac{-2+4}{2}\right)=(4,1)$. To get the radius, find the distance from the center to one endpoint: $\sqrt{(7-4)^2+(4-1)^2}=\sqrt{9+9}=\sqrt{18}$. Thus the equation is $(x-4)^2+(y-1)^2=18$, since the standard form uses $(x-h)^2+(y-k)^2=r^2$. The wrong choices show common traps. $(x-4)^2+(y-1)^2=72$ uses the squared length of the entire diameter — the distance between the endpoints is $\sqrt{72}$, and that is $4r^2$, not $r^2$. $(x-4)^2+(y-1)^2=36$ comes from dividing that squared diameter by 2 instead of 4, giving a radius of 6 rather than $\sqrt{18}$. Finally, $(x+4)^2+(y+1)^2=18$ has the correct radius but the center $(-4,-1)$; that sign error usually comes from miscomputing the midpoint. For the exam, follow a routine: midpoint for the center, then distance from the center to either endpoint for the radius. Always square that radius carefully — and never confuse it with the diameter.
12. A circle passes through $(-2,0)$ and $(6,0)$,and its center lies on the line $y=x$. Which equation represents this circle?
    Answer: $(x-2)^2+(y-2)^2=20$ When you see a circle question with two given points, think geometrically: the center must be equidistant from both points, so it lies on the perpendicular bisector of the segment joining them. Here the points are $(-2,0)$ and $(6,0)$, a horizontal segment with midpoint $(2,0)$, so the perpendicular bisector is $x=2$. Since the center also lies on $y=x$, the center must be $(2,2)$. The radius is the distance from $(2,2)$ to either point, say $(6,0)$: $\sqrt{(6-2)^2+(0-2)^2}=\sqrt{20}$. Thus the circle's equation is $(x-2)^2+(y-2)^2=20$. The choice $(x-2)^2+(y-2)^2=4$ has the correct center but radius $2$, which is too small to reach either given point. The choice $(x-2)^2+(y+2)^2=20$ has center $(2,-2)$, which is not on $y=x$; the plus sign in $y+2$ means the center's $y$-coordinate is $-2$. The choice $(x-2)^2+(y-2)^2=16$ uses radius $4$, a common trap: it treats the segment between $(-2,0)$ and $(6,0)$ as a diameter, but those are just two points on the circle, not opposite endpoints of a diameter. A strong strategy: when a circle passes through two points, immediately find the perpendicular bisector of those points. Its intersection with any other condition gives the center. Then calculate the radius by distance to one of the points. Never assume the given points are endpoints of a diameter unless the center is exactly their midpoint.
13. A parabola has focus at $(3,2)$and directrix $y=-4$. If a circle has the latus rectum of this parabola as its diameter, which equation represents the circle? (The latus rectum is the chord through the focus perpendicular to the parabola's axis of symmetry.)
    Answer: $(x-3)^2+(y-2)^2=36$ The vertex is midway between focus $(3,2)$and directrix $y=-4$, so it is $(3,-1)$,and $p=3$. The parabola opens upward, giving equation $(x-3)^2=12(y+1)$. On the latus rectum at $y=2$, $(x-3)^2=36$, so the endpoints are $(-3,2)$and $(9,2)$. The circle with that segment as diameter has center $(3,2)$and radius $6$, hence $(x-3)^2+(y-2)^2=36$. Radius squared $144$ results from treating the focus-to-directrix distance $6$ as $p$, which doubles the latus half-length; center $(3,-1)$ uses the vertex instead of the focus.
14. What is the circumference of the circle given by $x^2+y^2+6x-10y-15=0$?
    Answer: $14\pi$ Whenever you see a circle equation in expanded form, your first move is to complete the square to uncover the center and radius. For $x^2+y^2+6x-10y-15=0$, group the $x$ terms and the $y$ terms: $(x^2+6x)+(y^2-10y)=15$. Complete each square: $(x+3)^2-9+(y-5)^2-25=15$, so $(x+3)^2+(y-5)^2=49$. The radius is $\sqrt{49}=7$, and circumference is $2\pi r = 14\pi$. That is the correct choice. The distractor $7\pi$ is the radius itself, not the circumference — it forgets the factor of $2\pi$. The choice $49\pi$ is the area of the circle, $\pi r^2$, not its perimeter. The choice $98\pi$ is double the area, a sign of confusing area with circumference and then multiplying by 2. Each wrong answer comes from mixing up the circle formulas: radius, diameter, area, and circumference. On this exam, remember the key formulas: circumference is $2\pi r$ or $\pi d$, and area is $\pi r^2$. When the equation is given in expanded form, completing the square is always the first step — then read the radius straight from the standard form. A quick check: if the constant on the right is a perfect square like $49$, take its square root before doing anything with $\pi$.
15. A parabola has vertex $(2,-3)$and directrix $x=-1$. Which statement is correct?
    Answer: Its focus is $(5,-3)$,and its equation is $(y+3)^2=12(x-2)$. The directrix is to the left of the vertex, so the parabola opens to the right with $p=3$. The focus is $(2+3,-3)=(5,-3)$,and the equation is $(y+3)^2=4p(x-2)=12(x-2)$. The focus $(-1,-3)$ lies on the directrix and the negative-coefficient equation would open left; the focus $(2,0)$ and vertical-axis equation confuse the axis;and the equation $(y-3)^2=12(x+2)$ has sign errors in the vertex coordinates.
16. For what value of $k$ is the circle $x^2+y^2+6x-8y+k=0$ tangent to the y-axis?
    Answer: 16 Completing the square gives $(x+3)^2+(y-4)^2=25-k$, so the center is $(-3,4)$. Tangency to the y-axis means the radius is the distance from the center to the y-axis, which is $|-3|=3$. Thus $25-k=9$, so $k=16$. The value 9 would make it tangent to the x-axis, 25 gives a point circle, and 41 gives a negative radius squared.
17. A parabola is given by $y^2-6y+8x+1=0$. What are its focus and directrix?
    Answer: Focus $(-1,3)$; directrix $x=3$ When you see a parabola with the $y$ term squared, it opens left or right. Your first move is to complete the square to rewrite it in the standard form $(y-k)^2 = 4p(x-h)$. Here, $y^2-6y+8x+1=0$ becomes $y^2-6y = -8x-1$. Completing the square on the left: $(y-3)^2 - 9 = -8x -1$, so $(y-3)^2 = -8x +8 = -8(x-1)$. Thus $h=1$, $k=3$, and $4p=-8$ so $p=-2$. The focus is $(h+p,k) = (1-2,3) = (-1,3)$, and the directrix is $x = h-p = 1-(-2) = 3$. That matches the choice "Focus $(-1,3)$; directrix $x=3$". Now, why the others are traps. "Focus $(3,3)$; directrix $x=-1$" uses $p=+2$, which would come from $(y-3)^2 = 8(x-1)$ — the sign error of ignoring the negative coefficient. "Focus $(-3,3)$; directrix $x=5$" gives a vertex at $x=1$ but a distance $p=-4$ (or $+4$), double the correct magnitude, likely from misreading $4p=-8$ as $p=-4$. Finally, "Focus $(1,3)$; directrix $x=3$" confuses the vertex $(1,3)$ with the focus; the vertex is halfway between focus and directrix, so the focus would be $(-1,3)$ if the directrix is $x=3$. Your key check: after completing the square, compare $4p$ carefully — the sign tells you which way the parabola opens, and the distance $|p|$ sets the focus and directrix. Always compute $h+p$ and $h-p$ directly.
18. A parabola has vertex $(3,-2)$ and focus $(3,1)$. Which equation represents the parabola?
    Answer: $(x-3)^2=12(y+2)$ When you see a parabola with a vertex and focus, first determine which way it opens by comparing the coordinates. Here both have $x=3$, so the axis is vertical; since the focus $(3,1)$ is above the vertex $(3,-2)$, the parabola opens upward. For a vertical parabola, the standard form is $(x-h)^2=4p(y-k)$, where $(h,k)$ is the vertex and $p$ is the directed distance from vertex to focus. The distance from $-2$ to $1$ is $3$, so $p=3$, making $4p=12$. Substituting $h=3$, $k=-2$, you get: $$(x-3)^2=12(y+2)$$ That is the correct equation. Now look at the traps. $(x-3)^2=-12(y+2)$ would have $p=-3$, meaning the parabola opens downward — not what the focus indicates. $(y+2)^2=12(x-3)$ is a horizontal parabola opening right; its focus would be at $(6,-2)$, not $(3,1)$. Finally, $(x-3)^2=6(y+2)$ uses $4p=6$, so $p=1.5$, which is only half the actual vertex-to-focus distance. A good habit: always compute $p$ as the distance from vertex to focus, then multiply by $4$. Also match the squared variable to the axis — if the focus changes $y$, the parabola opens vertically, so $x$ is squared.
19. The circle $x^2+y^2-4x+8y+11=0$ is graphed. Which point lies outside the circle?
    Answer: $(5,-3)$ Whenever you see a circle equation like this, your first move is to rewrite it in standard form by completing the square — that reveals the center and radius, which are the keys to deciding whether a point is inside, on, or outside the circle. Complete the square: $x^2-4x + y^2+8y = -11$ $(x-2)^2 -4 + (y+4)^2 -16 = -11$ $(x-2)^2 + (y+4)^2 = 9$ So the center is $(2,-4)$ and the radius is $3$. A point lies outside if its distance from the center is greater than $3$. Check each candidate. - $(2,-1)$: distance from $(2,-4)$ is $|3| = 3$ — it lies on the circle, not outside. - $(4,-4)$: distance is $\sqrt{(2)^2+0^2}=2$ — inside. - $(5,-3)$: distance is $\sqrt{(3)^2+(1)^2}=\sqrt{10}\approx 3.16$ — greater than $3$, so this point is outside. - $(0,-2)$: distance is $\sqrt{(-2)^2+(2)^2}=\sqrt{8}\approx 2.83$ — inside. The trap here is confusing "on" with "outside" — a point exactly at distance $3$ is not outside. Also, make sure you complete the square correctly; the center is easy to misread if you forget to adjust the constant. Study tip: For any circle, always find the center and radius first. Then test points by computing their distance from the center — no need to plug into the original equation. Memorize the distance comparison: $<r$ inside, $=r$ on, $>r$ outside.
20. Consider the parabola $y^2-6y+8x+17=0$. Which statement is false?
    Answer: Its directrix is $x=-1$. Completing the square in $y$ gives $(y-3)^2=-8(x+1)$. Thus the vertex is $(-1,3)$, $p=-2$, and the parabola opens left with focus $(-3,3)$. The directrix is $x=h-p=-1-(-2)=1$, not $x=-1$. Therefore the false statement is the one claiming the directrix is $x=-1$.

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