SAT Circle Questions
Circle Challenge
24 original SAT-style questions covering the most important circle problems in the coordinate plane.
Standard form • Distance • Graphs • General form • Completing the square • SAT shortcuts
How to use this challenge
These are original SAT-style questions designed around the major circle question patterns students encounter in coordinate geometry. They are not reproduced College Board questions.
Try each question before opening the solution. The goal is not just to get the answer — it is to recognize which circle strategy the SAT is asking you to use.
Read the Center and Radius from Standard Form
These questions test whether you can immediately recognize the center \((h,k)\) and radius \(r\) from \((x-h)^2+(y-k)^2=r^2\). Watch the signs inside the parentheses.
- A \((4,3)\), radius \(5\)
- B \((4,-3)\), radius \(5\)
- C \((-4,3)\), radius \(25\)
- D \((-4,-3)\), radius \(25\)
Step 1: Match the equation
Compare with \[ (x-h)^2+(y-k)^2=r^2. \]
Step 2: Read the center
\(x-4\) gives \(h=4\). \(y+3=y-(-3)\), so \(k=-3\).
Step 3: Find the radius
\[ r=\sqrt{25}=5. \]
- A \(3\)
- B \(9\)
- C \(18\)
- D \(81\)
The number on the right is \(r^2\).
Therefore:
Write the Circle’s Standard Equation from Its Center and Radius
Given the center and radius, put them directly into standard form. The main challenge is handling negative coordinates correctly.
- A \((x+2)^2+(y-5)^2=36\)
- B \((x-2)^2+(y+5)^2=36\)
- C \((x-2)^2+(y-5)^2=6\)
- D \((x+2)^2+(y+5)^2=6\)
Start with: \[ (x-h)^2+(y-k)^2=r^2. \]
Substitute \(h=2\), \(k=-5\), and \(r=6\).
- A \((x+3)^2+(y-4)^2=6\)
- B \((x-3)^2+(y+4)^2=36\)
- C \((x+3)^2+(y-4)^2=36\)
- D \((x-3)^2+(y-4)^2=36\)
The radius is \(6\), so \(r^2=36\).
The center is \((-3,4)\). Therefore:
Find the Circle’s Equation from Its Center and a Point on the Circle
First find the distance from the center to the given point. That distance is the radius. Then substitute the center and radius into standard form.
- A \((x-1)^2+(y-2)^2=5\)
- B \((x-1)^2+(y-2)^2=25\)
- C \((x-5)^2+(y-5)^2=25\)
- D \((x+1)^2+(y+2)^2=25\)
Step 1: Find the radius
From \((1,2)\) to \((5,5)\), the horizontal change is \(4\), and the vertical change is \(3\).
Step 2: Write the equation
- A \(6\)
- B \(8\)
- C \(10\)
- D \(14\)
The horizontal difference is \(6\). The vertical difference is \(8\).
Find the Radius from the Center and a Point
Remember the geometric definition: every point on a circle is exactly one radius away from the center.
- A \(4\)
- B \(6\)
- C \(8\)
- D \(10\)
Both points have the same \(y\)-coordinate. Therefore the distance is horizontal.
- A \(8\)
- B \(9\)
- C \(10\)
- D \(12\)
The changes are \(6\) horizontally and \(8\) vertically.
Find a Missing Coordinate of a Point on a Circle
These problems look more mysterious, but the idea is simple: substitute the known coordinate into the circle equation and solve for the missing value.
- A \(-2\)
- B \(0\)
- C \(4\)
- D \(-3\)
Substitute \(x=7\).
Therefore: \[ k-2=\pm5. \]
Substitute \(y=6\).
Thus:
So \(a=4\) or \(a=-6\). The positive value is \(4\).
Determine Whether a Point Lies on, Inside, or Outside the Circle
Substitute the point into the left side of the equation. Compare the result with \(r^2\).
- A \((5,3)\)
- B \((2,4)\)
- C \((7,1)\)
- D \((4,1)\)
Test A:
This equals the right side, so the point is on the circle.
- A On the circle
- B Inside the circle
- C Outside the circle
- D At the center
The radius squared is \(36\). Since \(25<36\), the point is closer to the center than the radius.
Find the Center and Radius from General Form
When you see \(x^2+y^2+Dx+Ey+F=0\), identify the \(x\)- and \(y\)-coefficients. You can use completing the square or the center shortcut.
Here \(D=-12\) and \(E=10\).
Therefore: \[ h+k=-4+7=3. \]
Find the Radius Directly from General Form
Once you know the shortcut, the radius can be found without fully rewriting the equation into standard form.
SAT shortcut
Then take the square root to obtain \(r\).
Here: \[ D=-10,\quad E=6,\quad F=-15. \]
Here: \[ D=4,\quad E=-12,\quad F=20. \]
Convert General Form to Standard Form by Completing the Square
This is the long way — but it is important to understand it. Divide each linear coefficient by \(2\), square it, and add the resulting value to both sides.
- A \((x-4)^2+(y+3)^2=36\)
- B \((x+4)^2+(y-3)^2=36\)
- C \((x-4)^2+(y+3)^2=11\)
- D \((x-8)^2+(y+6)^2=36\)
For \(x^2-8x\): \[ \frac{-8}{2}=-4, \] so add \(16\).
For \(y^2+6y\): \[ \frac62=3, \] so add \(9\).
For \(x^2+10x\), add \[ \left(\frac{10}{2}\right)^2=25. \]
For \(y^2-4y\), add \[ \left(\frac{-4}{2}\right)^2=4. \]
Therefore:
Read the Center, Radius, and Equation from a Circle’s Graph
Graph questions require you to translate visual information into algebra. Find the center first, then determine the radius.
- A \((x+2)^2+(y-3)^2=6\)
- B \((x-2)^2+(y+3)^2=36\)
- C \((x+2)^2+(y-3)^2=36\)
- D \((x+2)^2+(y+3)^2=36\)
The center is \((-2,3)\). The horizontal distance from \(-2\) to \(4\) is \(6\). Therefore \(r=6\).
The leftmost point is \(x=1\), and the rightmost point is \(x=9\).
The diameter is therefore:
The radius is half the diameter:
Find the Center and Radius When the Diameter Endpoints Are Given
The midpoint of a diameter is the center. The radius is half the diameter.
- A \((4,4)\)
- B \((6,4)\)
- C \((8,4)\)
- D \((6,8)\)
The center is the midpoint:
- A \(4\)
- B \(5\)
- C \(6\)
- D \(10\)
The coordinate changes are \(8\) and \(6\).
The radius is half the diameter:
Understand How Changes in the Equation Move or Resize the Circle
Changes to \(h\), \(k\), or \(r\) change the graph. Think of the standard equation as a map of the circle.
- A It moves 5 units left.
- B It moves 5 units right.
- C It moves 5 units up.
- D Its radius increases by 5.
Original center: \[ (2,-3). \]
New center: \[ (7,-3). \]
The \(x\)-coordinate increased by \(5\). The \(y\)-coordinate and radius stayed the same.
- A \((x+4)^2+(y-1)^2=12\)
- B \((x+4)^2+(y-1)^2=18\)
- C \((x+4)^2+(y-1)^2=36\)
- D \((x+8)^2+(y-2)^2=36\)
The original equation has \[ r^2=9, \] so \(r=3\).
Twice the radius means: \[ r_{\text{new}}=6. \]
But the equation contains \(r^2\), so we need:
The center does not change.
🎯 The SAT Circle Decision Strategy
Before doing algebra, ask yourself one question: What is the problem actually asking me to find?
Need the center? Look at standard form or use \[ \left(-\frac D2,-\frac E2\right). \]
Need the radius? Find \(r^2\), then take the square root.
Given a point? Find its distance from the center.
Given a diameter? Think midpoint + half the diameter.
Given a graph? Read the center and radius visually before writing anything.
Ready for the next circle?
You have now practiced the major coordinate-plane circle patterns: standard form, distance, graphs, general form, completing the square, shortcuts, diameter, and transformations.
← Review the Circles LessonCircles in the Coordinate Plane
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