Circle Equations - Math 2

1. A circle has the property that it passes through points $(-2, 1)$, $(4, 1)$, and $(1, 4)$. If this circle also passes through point $(1, y_0)$ for some value $y_0 \neq 4$, what is $y_0$?
   Answer: $-2$ First, find the equation of the circle passing through $(-2,1)$, $(4,1)$, and $(1,4)$. Since the first two points have the same $y$-coordinate, the center lies on the perpendicular bisector of the segment connecting them, which is the vertical line $x = \frac{-2+4}{2} = 1$. So the center is at $(1, k)$ for some $k$. Using the fact that the center is equidistant from all three points: distance from $(1,k)$ to $(-2,1)$ equals distance from $(1,k)$ to $(1,4)$. We have $\sqrt{(-2-1)^2 + (1-k)^2} = \sqrt{(1-1)^2 + (4-k)^2}$, so $\sqrt{9 + (1-k)^2} = |4-k|$. Squaring both sides: $9 + (1-k)^2 = (4-k)^2$, so $9 + 1 - 2k + k^2 = 16 - 8k + k^2$. Simplifying: $10 - 2k = 16 - 8k$, so $6k = 6$ and $k = 1$. The center is $(1,1)$ and the radius is $\sqrt{(1-1)^2 + (4-1)^2} = 3$. The circle equation is $(x-1)^2 + (y-1)^2 = 9$. For point $(1, y_0)$ to be on this circle: $(1-1)^2 + (y_0-1)^2 = 9$, so $(y_0-1)^2 = 9$, giving $y_0 - 1 = \pm 3$. Thus $y_0 = 4$ or $y_0 = -2$. Since we're told $y_0 \neq 4$, we have $y_0 = -2$.
2. Consider the family of circles with equation $x^2 + y^2 - 4x + 2y + c = 0$, where $c$ is a parameter. For what range of values of $c$ do these equations represent actual circles (not points or empty sets)?
   Answer: $c < 5$ To determine when the equation represents a circle, complete the square: $x^2 - 4x + y^2 + 2y + c = 0$. Adding and subtracting the necessary constants: $(x^2 - 4x + 4) + (y^2 + 2y + 1) + c - 4 - 1 = 0$, which gives $(x-2)^2 + (y+1)^2 = 5 - c$. For this to represent a circle (not a point or empty set), we need the right side to be positive: $5 - c > 0$, so $c < 5$. When $c = 5$, the equation represents a single point (degenerate circle). When $c > 5$, the equation has no real solutions (empty set). Choice C and D incorrectly include the boundary case $c = 5$, while choice B uses the wrong boundary value.
3. A circle is inscribed in the square with vertices at $(0,0)$, $(6,0)$, $(6,6)$, and $(0,6)$. Point $Q$ is chosen randomly on the circle. What is the probability that both coordinates of $Q$ are greater than $3$?
   Answer: $\frac{1}{4}$ When you encounter probability questions involving geometric regions, you need to find the ratio of favorable area to total area. First, let's establish the setup. The square has side length 6, so the inscribed circle has radius 3 and center at $(3,3)$. The circle's equation is $(x-3)^2 + (y-3)^2 = 9$. You want the probability that both coordinates of point $Q$ are greater than 3. This means $Q$ must lie in the region where $x > 3$ and $y > 3$, which corresponds to the first quadrant relative to the circle's center. Since the circle is perfectly symmetric about both the vertical line $x = 3$ and horizontal line $y = 3$, this favorable region represents exactly one-quarter of the circle. The probability is therefore $\frac{\text{area of quarter-circle}}{\text{area of full circle}} = \frac{1}{4}$, making D correct. Let's examine why the other answers are wrong. Choice A ($\frac{1}{8}$) might tempt you if you mistakenly thought about the intersection of $x > 3$ and $y > 3$ as being more restrictive than it actually is. Choice B ($\frac{1}{6}$) doesn't correspond to any natural geometric division of the circle. Choice C ($\frac{1}{3}$) could result from incorrectly dividing the circle into three parts instead of recognizing the four-way symmetry. Remember: inscribed circles in squares create natural symmetries. When dealing with coordinate conditions like "both greater than the center," you're typically looking at one of four equal quadrants, giving probability $\frac{1}{4}$.
4. Circle $C_1$ has center $(2, -1)$ and radius $5$. Circle $C_2$ has equation $(x-2)^2 + (y+1)^2 = 9$. Point $M$ lies on both circles. What is the distance from $M$ to the center of $C_1$?
   Answer: $5$ Both circles have the same center $(2, -1)$. Circle $C_1$ has radius $5$ and circle $C_2$ has radius $3$. Since $M$ lies on $C_1$, the distance from $M$ to the center must equal the radius of $C_1$, which is $5$. Choice A gives the radius of $C_2$. Choice B is the difference of the radii. Choice D is the sum of the radii.
5. Points $M(2, 6)$ and $N(8, 2)$ are endpoints of a diameter of a circle. Point $P(a, b)$ lies on this circle such that the area of triangle $MPN$ is $12$. How many possible positions are there for point $P$?
   Answer: Exactly two such points $P$ exist on the circle When you encounter a problem involving a circle with given diameter endpoints and need to find points with specific geometric properties, start by using the key circle theorems and coordinate geometry principles. Since $M(2, 6)$ and $N(8, 2)$ are diameter endpoints, the center is at their midpoint: $(\frac{2+8}{2}, \frac{6+2}{2}) = (5, 4)$. The radius is half the distance $MN$: $r = \frac{1}{2}\sqrt{(8-2)^2 + (2-6)^2} = \frac{1}{2}\sqrt{52} = \sqrt{13}$. For any point $P$ on the circle, the area of triangle $MPN$ equals $\frac{1}{2} \times \text{base} \times \text{height}$. Using $MN$ as the base, we have base length $\sqrt{52} = 2\sqrt{13}$. For area = 12: $12 = \frac{1}{2} \times 2\sqrt{13} \times h$, so the height $h = \frac{12}{\sqrt{13}}$. The height represents the perpendicular distance from $P$ to line $MN$. Points at this exact distance from line $MN$ form two parallel lines, one on each side of $MN$. Since the required distance $\frac{12}{\sqrt{13}} < \sqrt{13}$ (the radius), both parallel lines intersect the circle, giving exactly two intersection points. Choice A is wrong because the required distance is less than the radius, so intersections exist. Choice B incorrectly assumes only one solution, missing that there are two sides to line $MN$. Choice C suggests four points, which would require two distances, but we have only one specific area requirement. Remember: when finding points on a circle at a fixed distance from a chord, there are typically two solutions—one on each side of the chord.
6. The equation $x^2 + y^2 + Dx + Ey + F = 0$ represents a circle with center $(-3, 4)$ and radius $7$. What is the value of $D + E + F$?
   Answer: $-26$ The standard form is $(x+3)^2 + (y-4)^2 = 49$. Expanding: $x^2 + 6x + 9 + y^2 - 8y + 16 = 49$, so $x^2 + y^2 + 6x - 8y - 24 = 0$. Thus $D = 6$, $E = -8$, $F = -24$, and $D + E + F = 6 + (-8) + (-24) = -26$. Choice A omits $D + E$. Choice C gets the sign wrong. Choice D gets both the calculation and sign wrong.
7. A circle has equation $x^2 + y^2 - 8x + 6y - 11 = 0$. Point $P$ is on the circle such that the $x$-coordinate of $P$ is twice its $y$-coordinate. How many such points $P$ exist?
   Answer: Exactly two such points exist on the circle First, rewrite the circle equation in standard form by completing the square: $x^2 - 8x + y^2 + 6y = 11$. Adding $(8/2)^2 = 16$ and $(6/2)^2 = 9$: $(x-4)^2 + (y+3)^2 = 11 + 16 + 9 = 36$. The circle has center $(4, -3)$ and radius $6$. For point $P(x,y)$ where $x = 2y$, substitute into the circle equation: $(2y-4)^2 + (y+3)^2 = 36$. Expanding: $4y^2 - 16y + 16 + y^2 + 6y + 9 = 36$, so $5y^2 - 10y + 25 = 36$, giving $5y^2 - 10y - 11 = 0$. Using the quadratic formula: $y = \frac{10 \pm \sqrt{100 + 220}}{10} = \frac{10 \pm \sqrt{320}}{10} = \frac{10 \pm 8\sqrt{5}}{10} = 1 \pm \frac{4\sqrt{5}}{5}$. Since the discriminant is positive, there are exactly two real solutions, meaning exactly two points on the circle satisfy the condition.
8. Two circles have equations $x^2 + y^2 - 6x + 4y - 12 = 0$ and $x^2 + y^2 - 10x - 6y + 18 = 0$. At how many points do these circles intersect?
   Answer: $2$ Convert to standard form. Circle 1: $(x-3)^2 + (y+2)^2 = 25$ with center $(3, -2)$ and radius $5$. Circle 2: $(x-5)^2 + (y-3)^2 = 16$ with center $(5, 3)$ and radius $4$. The distance between centers is $\sqrt{(5-3)^2 + (3-(-2))^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.39$. Since $|5-4| = 1 < \sqrt{29} < 5+4 = 9$, the circles intersect at exactly $2$ points. Choice A would occur if circles don't intersect. Choice B would occur if circles are tangent. Choice D suggests insufficient information.
9. Circle $A$ has equation $(x-3)^2 + (y+2)^2 = 16$ and circle $B$ has equation $(x+1)^2 + (y-1)^2 = 9$. Point $T$ is equidistant from the centers of both circles. If $T$ lies on circle $A$, what is the $y$-coordinate of $T$?
   Answer: $2$ Circle $A$ has center $(3, -2)$ and circle $B$ has center $(-1, 1)$. Point $T$ equidistant from both centers lies on the perpendicular bisector of the segment connecting the centers. The midpoint is $(1, -\frac{1}{2})$ and the slope between centers is $\frac{1-(-2)}{-1-3} = -\frac{3}{4}$. The perpendicular bisector has slope $\frac{4}{3}$ and equation $y + \frac{1}{2} = \frac{4}{3}(x-1)$. Since $T$ is on circle $A$: $(x-3)^2 + (y+2)^2 = 16$. Solving simultaneously gives $T = (3, 2)$ or $T = (-1, -6)$. Only $(3, 2)$ gives $y = 2$.
10. A circle has equation $(x-3)^2 + (y+1)^2 = 25$. Line $\ell$ has equation $4x - 3y = k$ for some constant $k$. For how many values of $k$ does line $\ell$ intersect the circle at exactly one point?
    Answer: Exactly two values of $k$ result in exactly one intersection A line intersects a circle at exactly one point when the line is tangent to the circle. This occurs when the distance from the circle's center to the line equals the radius. The circle has center $(3, -1)$ and radius $5$. The distance from point $(3, -1)$ to line $4x - 3y - k = 0$ is $\frac{|4(3) - 3(-1) - k|}{\sqrt{4^2 + (-3)^2}} = \frac{|12 + 3 - k|}{5} = \frac{|15 - k|}{5}$. For tangency, this distance must equal the radius: $\frac{|15 - k|}{5} = 5$, so $|15 - k| = 25$. This gives $15 - k = 25$ or $15 - k = -25$, so $k = -10$ or $k = 40$. Therefore, there are exactly two values of $k$ that result in exactly one intersection point.
11. The equation $x^2 + y^2 + 6x - 8y + k = 0$ represents a circle that passes through the origin. What is the area of this circle?
    Answer: $25\pi$ Since the circle passes through the origin $(0,0)$, substituting gives $0 + 0 + 0 - 0 + k = 0$, so $k = 0$. The equation becomes $x^2 + y^2 + 6x - 8y = 0$. Completing the square: $(x+3)^2 - 9 + (y-4)^2 - 16 = 0$, so $(x+3)^2 + (y-4)^2 = 25$. The radius is $5$ and the area is $\pi r^2 = 25\pi$. Choice B doubles the area. Choice C uses radius $1$. Choice D uses diameter squared instead of radius squared.

Start this quiz