AP Calculus & Pre-Calc · Stewart 9e Diagnostic
Equation of a Circle from Center and a Point (Stewart Calculus 9e Q2)
Stewart's Calculus 9th Edition diagnostic Question 2 hands you a center and one point that the circle has to pass through, and asks for the equation. The whole problem is two ideas glued together — the standard form of a circle, and the distance formula. Done in the right order, it's a sixty-second problem. Done in the wrong order, students take an unnecessary square root and then square it again, lose a sign somewhere, and walk away with a radius they don't trust. Here is the SOMATH way.
Direct answer. The circle in standard form is (x − h)2 + (y − k)2 = r2, where (h, k) is the center. With center (−1, 4) we have h = −1 and k = 4. To get r2, use the squared distance from the center to the point on the circle: r2 = (3 − (−1))2 + (−2 − 4)2 = 16 + 36 = 52. So the equation is (x + 1)2 + (y − 4)2 = 52, with radius r = √52 = 2√13 ≈ 7.21.
Why Stewart wants this problem on the diagnostic
Stewart's Q2 is the constructive twin of Q3. Q3 hands you a circle in general form and asks you to recover its center and radius — a deconstruction problem. Q2 hands you the pieces and asks you to assemble the equation — a construction problem. Both lean on the same template, (x − h)2 + (y − k)2 = r2, but they pull on different reflexes. Q3 tests whether you can complete the square. Q2 tests whether you can read coordinates and apply the distance formula without overthinking it.
The reason Stewart drops this problem at the beginning of the diagnostic is that the distance formula is everywhere in calculus. It shows up in arc length, in the limit definition of magnitude for vectors, in the geometry of optimization problems where you minimize distance to a curve, and as the kernel of the dot product. If a student is fuzzy on whether the distance squared is (Δx)2 + (Δy)2 or needs a square root, calculus problems start to feel arbitrary. Stewart is making sure the foundation is firm before the calculus chapters arrive.
Step 1: write down the standard form
Don't reach for a formula sheet. The standard form of a circle should be reflexive by the time you sit for AP Calculus:
(x − h)2 + (y − k)2 = r2
The pair (h, k) is the center. The right-hand side is r2 — the radius squared, not the radius. That distinction matters because we are about to compute r2 directly from the distance formula and we do not need to take a square root to write the equation.
Step 2: plug in the center
Stewart gives us center (−1, 4). Substitute h = −1 and k = 4 into the template:
(x − (−1))2 + (y − 4)2 = r2
Clean up the double-negative in the first parenthesis. Subtracting −1 is the same as adding 1:
(x + 1)2 + (y − 4)2 = r2
The equation now describes every circle centered at (−1, 4), of any radius. We still need the size, which is where the point (3, −2) comes in.
Step 3: use the distance formula to get r²
The radius is the distance from the center to any point on the circle. Stewart gives us one such point — (3, −2). The distance formula in two dimensions is:
d = √[(x2 − x1)2 + (y2 − y1)2]
But we don't actually want d — we want d2, because that is what plugs straight into the equation. Square both sides of the distance formula and the square root disappears:
d2 = (x2 − x1)2 + (y2 − y1)2
That is the move worth memorizing. The unsquared distance formula has a square root in it; the squared distance has no square root. When the problem hands you a circle, you want the squared form because the equation of a circle is built on r2. Skipping the square-root-and-then-square-again roundtrip saves time and prevents arithmetic errors.
Substitute the center (−1, 4) as (x1, y1) and the point on the circle (3, −2) as (x2, y2):
r2 = (3 − (−1))2 + (−2 − 4)2
Simplify each parenthesis carefully. 3 − (−1) = 3 + 1 = 4, and −2 − 4 = −6:
r2 = 42 + (−6)2 = 16 + 36 = 52
So r2 = 52. Note that we never took a square root. We don't need to; the equation of the circle calls for r2 directly.
Step 4: write the final equation
Substitute r2 = 52 back into the equation we set up in Step 2:
(x + 1)2 + (y − 4)2 = 52
That is the answer Stewart is looking for. It tells you everything: center at (−1, 4), and the circle reaches out far enough to touch (3, −2). If a problem asks for the radius numerically, take the square root once at the end: r = √52 = √(4 · 13) = 2√13 ≈ 7.21. The factored radical form 2√13 is what graders prefer because it's exact.
The three mistakes worth memorizing
- Sign-flipping the center inside the parenthesis. Standard form is (x − h)2. With h = −1, the parenthesis becomes (x − (−1))2 = (x + 1)2. Students who write (x − 1)2 have moved the center to (+1, 4) and the rest of the answer comes out wrong. Always read the plus sign inside a squared binomial as "minus negative."
- Taking an unnecessary square root. The equation of a circle uses r2, not r. Compute r2 straight from the squared distance formula and plug it in. If you take the square root to find r = √52 and then square it again, you'll do the work twice and probably make an arithmetic slip. Skip that step.
- Confusing the two points. The distance formula gives the distance between two specific points: the center and the point on the circle. It does not matter which one you label (x1, y1) because the differences are squared — but it does matter that you keep both coordinates of each point together. Mixing the x of one point with the y of another flips the geometry. Stack the two points vertically on scratch paper before you subtract.
How SOMATH teaches this problem
At SOMATH we treat Q2 and Q3 together. Q2 builds the equation forward — center plus a point gives you r2 and you fill in the template. Q3 disassembles a general-form equation backward by completing the square. We drill them in pairs because students who can do both fluently never get confused about which version of the equation they are looking at. The standard form is the bridge: every circle problem on the AP Calc exam, the SHSAT, the SAT, and the in-class assessments at the New York City independent schools we serve maps to one of these two operations.
Our classroom on 226 W 79th St (Upper West Side, between Broadway and Amsterdam) runs small groups of three to five students with whiteboards on every wall. We don't hand out worksheet stacks. Each technique gets introduced with the why, then practiced on three increasingly hostile problems, then tied back to the calculus context where the student will meet it next. Stewart's Q2 takes ninety seconds in class; the lesson around it — the distance formula, the squared shortcut, the sign convention inside the parentheses, and the connection to vector magnitudes — takes about thirty minutes and pays off for the next two years.
Want your child in an AP Pre-Calculus class at SOMATH?
Full-year course covering polynomial, rational, exponential, logarithmic, and trigonometric functions plus sequences and series. Delivered with the depth needed to walk into AP Calculus prepared.
FAQ
What is the standard form of a circle?
Standard form is (x − h)2 + (y − k)2 = r2, where (h, k) is the center and r is the radius. The right-hand side is r2, not r — a distinction that matters for both Q2 (where you compute r2 directly from a point) and Q3 (where you complete the square to recover it).
Do I always need the distance formula on this kind of problem?
Yes, in some form. If you are given the center and a point on the circle, the squared distance between them is r2. You can write that out as the distance formula, or you can simply subtract the coordinates and square — same arithmetic, just less notation. Either way, the geometric idea is identical: the radius is the distance from the center to any point on the circle.
How do you simplify sqrt(52)?
Factor under the radical to look for perfect squares: 52 = 4 · 13, and 4 is a perfect square. So √52 = √(4 · 13) = √4 · √13 = 2√13. Numerically, 2√13 ≈ 2 · 3.606 ≈ 7.21. On exams that grade for exact answers, write 2√13. On problems that ask for a decimal, write 7.21.
Will I see this exact problem type in AP Calculus?
The exact form rarely appears in AP Calc multiple choice, but the underlying skill — applying the distance formula in squared form, manipulating (x − h)2-style templates, and recognizing conics from general or standard form — is embedded throughout free-response geometry-of-curves problems, parametric and polar conversions, and optimization. The Stewart diagnostic is intentionally upstream of all of that.
What about Q1?
Q1 of Stewart's analytic-geometry diagnostic asks for the equation of a line given a slope and a point — same idea (template plus given data), different geometric object. We have a full walkthrough of that one too, linked at the bottom of this post.
Free 30-minute evaluation
Every new SOMATH student starts with a free 30-minute evaluation and a written diagnostic delivered within 48 hours — even if you don't enroll. AP Calculus, college calculus, pre-calc, SHSAT, SAT, and everything in between. 226 W 79th St, Upper West Side.