1 which point lies on the circle represented by the equation ((x - 1)^2 + (y + 2)^2 = 25)?
○ a ((0, 6))
○ b ((0, -7))
○ c ((1, -2))
○ d ((3, 1))

2 a tree casts a shadow of 15 meters. a 2 - meter stick casts a shadow of 3 meters. what is the height of the tree?
○ a 12 meters
○ b 10 meters
○ c 7.5 meters
○ d 5 meters

3 which point is equidistant from all vertices of a triangle?
○ a centroid
○ b circumcenter
○ c tangent
○ d secant

4 what role does the perpendicular bisector play in constructing a tangent to a circle from an external point?
○ a it intersects the circle at only one point.
○ b it passes through the center of the circle but does not intersect the circle.
○ c it helps us find the midpoint of the line connecting the center of the circle to the external point. the midpoint becomes the center of a new circle, and the points where the new circle intersects the original circle are the points of tangency.
○ d it is parallel to the tangent line.

5 what real - world object’s shape most closely matches a plane?
○ a a basketball
○ b a marble
○ c a pencil
○ d a tabletop

6 which of the following is the correct general form of the equation of a circle with center ((2, -3)) and radius 3?
○ a (x^2 + y^2 - 4x + 6y + 4 = 0)
○ b (x^2 + y^2 + 4x - 6y + 16 = 0)
○ c (x^2 + y^2 + 4x + 6y - 4 = 0)
○ d (x^2 + y^2 - 4x + 6y - 12 = 0)

7 what is the geometric term for a flat surface that extends without end in all directions?
○ a line segment
○ b angle
○ c plane
○ d ray

Question

1 which point lies on the circle represented by the equation ((x - 1)^2 + (y + 2)^2 = 25)?
○ a ((0, 6))
○ b ((0, -7))
○ c ((1, -2))
○ d ((3, 1))

2 a tree casts a shadow of 15 meters. a 2 - meter stick casts a shadow of 3 meters. what is the height of the tree?
○ a 12 meters
○ b 10 meters
○ c 7.5 meters
○ d 5 meters

3 which point is equidistant from all vertices of a triangle?
○ a centroid
○ b circumcenter
○ c tangent
○ d secant

4 what role does the perpendicular bisector play in constructing a tangent to a circle from an external point?
○ a it intersects the circle at only one point.
○ b it passes through the center of the circle but does not intersect the circle.
○ c it helps us find the midpoint of the line connecting the center of the circle to the external point. the midpoint becomes the center of a new circle, and the points where the new circle intersects the original circle are the points of tangency.
○ d it is parallel to the tangent line.

5 what real - world object’s shape most closely matches a plane?
○ a a basketball
○ b a marble
○ c a pencil
○ d a tabletop

6 which of the following is the correct general form of the equation of a circle with center ((2, -3)) and radius 3?
○ a (x^2 + y^2 - 4x + 6y + 4 = 0)
○ b (x^2 + y^2 + 4x - 6y + 16 = 0)
○ c (x^2 + y^2 + 4x + 6y - 4 = 0)
○ d (x^2 + y^2 - 4x + 6y - 12 = 0)

7 what is the geometric term for a flat surface that extends without end in all directions?
○ a line segment
○ b angle
○ c plane
○ d ray

Accepted Answer

### Question 1
# Explanation:
## Step1: Recall circle equation form
The standard form of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. Here, center is $(1, - 2)$ and $r = 5$ (since $r^2=25$).
## Step2: Check each point
- For point $(6, - 2)$: Substitute into $(x - 1)^2+(y + 2)^2$. $(6 - 1)^2+(-2 + 2)^2=25+0 = 25$, which equals $r^2$. So this point is on the circle. (Let's check others briefly: For $(0, - 7)$: $(0 - 1)^2+(-7 + 2)^2=1 + 25 = 26
eq25$; For $(1, - 2)$: $(1 - 1)^2+(-2 + 2)^2=0
eq25$ (it's the center); For $(5,1)$: $(5 - 1)^2+(1 + 2)^2=16 + 9 = 25$? Wait, $(5 - 1)^2=16$, $(1 + 2)^2 = 9$, $16+9 = 25$. Wait, maybe I made a mistake earlier. Wait, $(6,-2)$: $(6 - 1)=5$, $5^2 = 25$, $(-2 + 2)=0$, so $25+0 = 25$. $(5,1)$: $(5 - 1)=4$, $4^2 = 16$, $(1 + 2)=3$, $3^2 = 9$, $16 + 9 = 25$. Wait, but maybe the original options were misread. Wait, the first option: if it's $(6, - 2)$, then it's on the circle. (Assuming the first option is $(6, - 2)$ as per typical problems)
# Answer: The point $(6, - 2)$ (assuming option c is $(6, - 2)$)

### Question 2
# Explanation:
## Step1: Set up proportion
Let the height of the tree be $h$. The ratio of height to shadow length should be equal (similar triangles). So $\frac{h}{15}=\frac{2}{3}$.
## Step2: Solve for $h$
Cross - multiply: $3h=15\times2$, so $3h = 30$, then $h = 10$ meters.
# Answer: B. 10 meters

### Question 3
# Explanation:
The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect, and it is equidistant from all three vertices of the triangle. The centroid is the intersection of medians (equidistant in a different sense), tangent is a line, and secant is a line.
# Answer: b. Circumcenter

### Question 4
# Explanation:
When constructing a tangent from an external point to a circle, the perpendicular bisector of the segment joining the external point to the center helps find the mid - point. This mid - point is the center of a new circle (with radius equal to half the distance between external point and center), and the intersection points of this new circle with the original circle are the points of tangency. Option a is wrong (perpendicular bisector isn't a tangent), option b is wrong (it connects external point and center, so it intersects the circle), option d is wrong (it's not parallel to tangent).
# Answer: c. It helps us find the midpoint of the line connecting the center of the circle to the external point. The midpoint becomes the center of a new circle, and the points where the new circle intersects the original circle are the points of tangency

### Question 5
# Explanation:
A plane is a flat, two - dimensional surface that extends infinitely. A tabletop is a flat surface that most closely resembles a plane (a basketball is a sphere, a marble is a sphere, a pencil is a cylinder).
# Answer: d. A tabletop

### Question 6
# Explanation:
## Step1: Recall general form of circle equation
The general form is $x^{2}+y^{2}+Dx + Ey+F = 0$, and the standard form is $(x - h)^2+(y - k)^2=r^2$. Given center $(2,-3)$ and $r = 1$, standard form is $(x - 2)^2+(y + 3)^2=1$.
## Step2: Expand the standard form
Expand $(x - 2)^2+(y + 3)^2=1$: $x^{2}-4x + 4+y^{2}+6y + 9 = 1$.
Simplify to general form: $x^{2}+y^{2}-4x + 6y+12 = 0$? Wait, no: $x^{2}-4x + 4+y^{2}+6y + 9-1 = 0\Rightarrow x^{2}+y^{2}-4x + 6y + 12 = 0$? Wait, let's recalculate. $(x - 2)^2=x^{2}-4x + 4$, $(y + 3)^2=y^{2}+6y + 9$. So $(x - 2)^2+(y + 3)^2=1\Rightarrow x^{2}-4x + 4+y^{2}+6y + 9 = 1\Rightarrow x^{2}+y^{2}-4x + 6y+12 = 0$. Wait, but let's check the options. Option a: $x^{2}+y^{2}-4x + 6y + 4 = 0$. Wait, maybe I made a mistake. Wait, $r = 1$, so $(x - 2)^2+(y + 3)^2=1\Rightarrow x^{2}-4x + 4+y^{2}+6y + 9 = 1\Rightarrow x^{2}+y^{2}-4x + 6y+12 = 0$. Wait, the options: option d is $x^{2}+y^{2}-4x + 6y-12 = 0$? No, wait, maybe the radius is 3? Wait, no, the question says radius 1. Wait, maybe a typo, but let's check the expansion again. Wait, $(x - 2)^2+(y + 3)^2=1$ expands to $x^{2}-4x + 4 + y^{2}+6y + 9=1$, so $x^{2}+y^{2}-4x + 6y+12 = 0$. But looking at the options, option a: $x^{2}+y^{2}-4x + 6y + 4 = 0$. Wait, maybe the radius is 3? If $r = 3$, then $(x - 2)^2+(y + 3)^2=9$, expand: $x^{2}-4x + 4+y^{2}+6y + 9 = 9\Rightarrow x^{2}+y^{2}-4x + 6y + 4 = 0$, which is option a. Maybe the radius was 3 (a common mistake). So assuming radius is 3, the correct option is a.
# Answer: a. $x^{2}+y^{2}-4x + 6y + 4 = 0$

### Question 7
# Explanation:
A plane is defined as a flat surface that extends without end in all directions. A line segment has endpoints, an angle is formed by two rays, a line extends in two directions but is one - dimensional. So the geometric term for a flat surface that extends without end in all directions is a plane.
# Answer: c. Plane

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Recall circle equation form

Step2: Check each point

Step1: Set up proportion

Step2: Solve for \(h\)

Answer:

Question 2