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Circles in the Coordinate Plane Practice Quiz

Improve skills with engaging coordinate circle exercises

Difficulty: Moderate
Grade: Grade 9
Study OutcomesCheat Sheet
Colorful paper art promoting a Coordinate Circle Challenge trivia quiz for high school students.

What is the standard form of a circle's equation?
x² + y² = r²
(x - h)² + (y - k)² = r²
(x - h)² - (y - k)² = r²
(x + h)² + (y + k)² = r²
The standard form of a circle's equation is (x - h)² + (y - k)² = r² where (h, k) represents the center and r is the radius. This form clearly shows the geometric features of the circle.
What does the center (h, k) represent in the circle equation (x - h)² + (y - k)² = r²?
A point on the circle
The slope of the circle
The coordinates of the center
The radius of the circle
In the equation, h and k designate the x and y coordinates of the circle's center. This directly identifies the location of the circle on the coordinate plane.
In the equation (x - 3)² + (y + 2)² = 16, what is the radius of the circle?
-4
8
16
4
The radius is found by taking the square root of 16, which results in 4. This follows directly from the standard form where r² is on the right side of the equation.
Which of the following points is the center of the circle given by (x + 1)² + (y - 5)² = 9?
(1, 5)
(-1, 5)
(1, -5)
(-1, -5)
The expression (x + 1)² can be rewritten as (x - (-1))², showing that the x-coordinate of the center is -1. Similarly, (y - 5)² indicates that the y-coordinate is 5, so the center is (-1, 5).
What must be true for a point (x, y) to lie on a circle defined by (x - h)² + (y - k)² = r²?
It must satisfy (x - h)² + (y - k)² < r²
It must satisfy the equation (x - h)² + (y - k)² = r²
It must satisfy (x - h)² - (y - k)² = r²
It must satisfy (x - h) + (y - k) = r²
A point lies on the circle if, when substituted into the equation, the equality holds true. This differentiates points on the circle from those inside or outside it.
How can you determine the center and radius from the standard form of a circle's equation?
By completing the square for the x and y terms
By identifying h and k as the center and taking the square root of r²
By setting the equation equal to zero and solving for x and y
By dividing the equation by r²
The standard form (x - h)² + (y - k)² = r² explicitly shows the center as (h, k) and the radius as the square root of the right-hand side. Completing the square is necessary only when converting from the general form.
If the equation of a circle is given by x² + y² + 6x - 4y + 9 = 0, what is the center of the circle after rewriting it in standard form?
(-3, -2)
(3, 2)
(3, -2)
(-3, 2)
Completing the square for both x and y transforms the equation into (x + 3)² + (y - 2)² = 4, which reveals the center as (-3, 2). This method is standard for converting the general form into the standard form.
What is the radius of a circle whose standard form equation is (x - 4)² + (y + 3)² = 25?
5
25
-5
10
Since the equation shows that r² = 25, the radius is the positive square root of 25, which is 5. This direct calculation confirms the circle's size.
For the circle (x + 2)² + (y - 6)² = 49, which of the following points lies on the circle?
(5, 6)
(2, 8)
(-1, -1)
(-4, 3)
Substituting (5, 6) into the equation yields (5 + 2)² + (6 - 6)² = 7² + 0² = 49, which satisfies the circle's equation. The other options do not meet this condition.
After expanding the standard form (x - h)² + (y - k)² = r², which term represents the sum of the x terms when written in the general form?
h x²
2h x
-2h x
-h² x
Expanding (x - h)² gives x² - 2hx + h², so the linear term in x is -2hx. This term connects the standard form to the general form of the circle's equation.
Which step is necessary to convert the general form of a circle's equation to its standard form?
Setting the discriminant equal to zero
Factoring out the common variable
Completing the square for both the x and y variables
Dividing the entire equation by the constant term
Converting from the general to the standard form requires completing the square for both x and y terms. This method restructures the equation to clearly display the center and radius.
What is the result of calculating the distance between the center (-2, 1) and a point on the circle (4, 5)?
√13
13
2√13
4√13
Using the distance formula, the distance is calculated as √[(4 - (-2))² + (5 - 1)²] = √(6² + 4²) = √(36 + 16) = √52, which simplifies to 2√13. This confirms the distance between the two points.
If a circle's radius is doubled, how does its area change?
It increases by a factor of 8
It doubles
It remains the same
It quadruples
Since the area of a circle is given by πr², doubling the radius results in an area of π(2r)² = 4πr². Thus, the area becomes four times as large.
Which of these equations represents a circle tangent to the x-axis?
(x - 1)² + (y - 3)² = 16
(x + 2)² + (y - 2)² = 1
(x - 1)² + (y - 3)² = 4
(x - 1)² + (y - 3)² = 9
A circle is tangent to the x-axis if the absolute value of the y-coordinate of its center equals the radius. For (x - 1)² + (y - 3)² = 9, the center is (1, 3) and the radius is 3, which satisfies this condition.
Which transformation will shift the circle (x - 2)² + (y - 2)² = 16 so that its center is moved to the origin?
Reflect across the line y = x
Rotate 90 degrees about the origin
Translate by (2, 2)
Translate by (-2, -2)
The circle's center is at (2, 2) and moving it to the origin requires subtracting 2 from both coordinates, which means a translation by (-2, -2). Other transformations do not achieve the required shift.
Given the circle with equation x² + y² + 8x - 10y + 5 = 0, determine its center and radius.
Center: (-4, 5), Radius: 36
Center: (4, -5), Radius: 6
Center: (-4, 5), Radius: 6
Center: (4, 5), Radius: 6
Completing the square for x and y transforms the equation into (x + 4)² + (y - 5)² = 36, revealing the center as (-4, 5) and the radius as √36, which is 6. This process is standard for rewriting a circle's equation into its standard form.
For a circle with center (h, k) and radius r, if a line given by y = mx + b is tangent to the circle, what condition must be satisfied?
The y-intercept b equals r
The distance from (h, k) to the line, |mh - k + b|/√(m²+1), equals r
The point (h, k) lies on the line
The line's slope m equals r
A line is tangent to a circle if the perpendicular distance from the circle's center to the line equals the radius. The distance formula |mh - k + b|/√(m²+1) set equal to r provides the necessary condition.
If the circle (x - a)² + (y - 3)² = 16 is tangent to the line y = x, what is one possible value for a?
3
4√2 - 3
3 + 4√2
3 - 4√2
The perpendicular distance from the center (a, 3) to the line y = x is |a - 3|/√2. Setting this equal to the radius 4 yields |a - 3| = 4√2, so one possible solution is a = 3 + 4√2. Although another solution exists, only one option is marked correct.
Determine the equation of a circle with center (-2, 4) that is externally tangent to the circle (x - 1)² + (y + 1)² = 25.
(x + 2)² + (y - 4)² = (√34 + 5)²
(x - 2)² + (y - 4)² = (√34 - 5)²
(x + 2)² + (y - 4)² = (√34 - 5)²
(x - 2)² + (y + 4)² = (√34 - 5)²
For externally tangent circles, the sum of the radii equals the distance between the centers. The distance between (-2, 4) and (1, -1) is √34, so the radius of the circle with center (-2, 4) is √34 - 5, leading to the equation (x + 2)² + (y - 4)² = (√34 - 5)².
A circle has its equation in general form: x² + y² + Dx + Ey + F = 0. Which condition must hold for this equation to represent a real circle with a positive radius?
(D/2)² + (E/2)² - F = 0
D + E + F > 0
F > (D/2)² + (E/2)²
(D/2)² + (E/2)² - F > 0
By completing the square, the general form converts to the standard form with the radius squared given by (D/2)² + (E/2)² - F. This value must be greater than zero for the circle to have a positive radius.
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Study Outcomes

  1. Identify the center and radius of circles from their equations.
  2. Convert general circle equations to standard form.
  3. Solve problems involving the positioning of circles in the coordinate plane.
  4. Evaluate the impact of changes in circle parameters on its graph.
  5. Verify solutions through graphing techniques and analytical methods.

Circles in Coordinate Plane Practice Cheat Sheet

  1. Standard Equation of a Circle - Every circle in coordinate geometry is defined by (x − h)2 + (y − k)2 = r2, where (h, k) is the center and r is the radius. This formula is your go‑to tool for graphing and analyzing circles with confidence. MathBits Notebook
  2. Deriving with the Pythagorean Theorem - Picture a right triangle formed from the center to any point on the circle; the Pythagorean Theorem then unveils the circle's equation. This approach connects algebra and geometry in a single aha moment. Geometry Common Core
  3. Completing the Square - Transform x² + y² + Dx + Ey + F = 0 into standard form by grouping x's and y's and adding perfect squares. This process reveals the center and radius hiding in the general equation. OpenAlgebra
  4. Graphing Circles - Start by plotting the center (h, k), then step out a distance r in all directions to sketch the circle. This hands‑on method cements the link between equation and shape. GeoGebra
  5. Identifying Center & Radius - In (x − 3)2 + (y + 2)2 = 16, the center is (3, −2) and r = 4. Spotting those values quickly speeds up problem solving. Leaving Cert Maths
  6. Point Location Tests - Plug any point into the circle's equation: if the result equals r² it's on the circle, less means inside, more means outside. This simple check is a student favorite. Leaving Cert Maths
  7. Real‑World Applications - From designing wheels to tracking planetary orbits, circle equations model countless real scenarios. Appreciating these examples brings textbook formulas to life. Dummies.com
  8. Equation from Diameter Endpoints - Find the midpoint of the endpoints for the center, compute half the distance as r, then plug into standard form. It's a neat coordinate geometry workout! MathBits Notebook
  9. Interactive Exploration - Use dynamic tools like GeoGebra to drag the center or tweak the radius and watch the circle update in real time. This interactive play boosts intuition. GeoGebra
  10. Common Pitfalls - Watch out for sign mix‑ups with (h, k) and forgetting that r is the radius, not its square. A little caution here saves major headaches later! Geometry Common Core
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