AP Calculus & Pre-Calc · Stewart 9e Diagnostic
Analytic Geometry Diagnostic Q3 (Stewart Calculus 9e): Center and Radius of a Circle
Stewart's Calculus 9th Edition diagnostic doesn't just check whether you remember the equation of a circle — Question 3 hands you the general form and asks you to recover its center and radius. The whole problem hinges on one technique that calculus students will reuse a hundred times: completing the square. We walk through it the way we teach it at SOMATH on the Upper West Side, with the sign traps, the half-square trick, and the final check that catches the most common mistake.
Direct answer. The equation x2 + y2 − 6x + 10y + 9 = 0 is the general form of a circle. Complete the square in both variables: add (6/2)2 = 9 for the x-group and (10/2)2 = 25 for the y-group, balancing both sides. The equation becomes (x − 3)2 + (y + 5)2 = 25. Compare with standard form (x − h)2 + (y − k)2 = r2 and read off the answers: center (3, −5) and radius 5. The whole problem is a single algebra move applied twice, with one final square root at the end.
Why Stewart wants this problem on the diagnostic
Completing the square is one of the few algebra techniques that shows up unchanged in calculus. It is how you derive the quadratic formula, how you integrate rational functions whose denominators are quadratics with no real roots, and how you finish off most conic-section problems before they ever reach the calculus chapter. If a student cannot complete the square fluently — meaning, can run the procedure without thinking through the why — limits of rational functions and arc-length integrals are going to feel artificially hard.
Stewart's Question 3 picks the cleanest possible test case. The coefficients on x2 and y2 are both 1, the linear coefficients are even integers, and the final answer comes out to a perfect-square radius. There is nothing exotic to navigate; the problem is purely a check on whether the procedure is reliable. At SOMATH we use this exact problem in placement sessions for the same reason — it's surgical.
Step 1: rearrange and group
Start with the equation given by Stewart:
x2 + y2 − 6x + 10y + 9 = 0
Move the constant to the right-hand side, then group the x-terms and y-terms together:
x2 − 6x + y2 + 10y = −9
That rearrangement does the structural work. Each variable now sits in its own quadratic that we will complete separately. Notice that the leading coefficient on each squared term is already 1 — that is what makes this version of the problem clean. If those coefficients were anything else, the first move would be to divide through. They aren't, so we don't.
Step 2: complete the square in x
Look at x2 − 6x. To make this a perfect square, take half of the linear coefficient and square it:
−62 = −3, and (−3)2 = 9
Adding 9 turns the x-group into a perfect square:
x2 − 6x + 9 = (x − 3)2
The −3 inside the parenthesis is exactly the same as the half of the linear coefficient we just calculated. That is the whole trick — half of −6 is −3, and that −3 becomes the constant inside the squared binomial. To keep the equation balanced, whatever we add on the left also has to appear on the right.
Step 3: complete the square in y
Same routine on the y-group y2 + 10y:
102 = 5, and 52 = 25
Adding 25 completes the square in y:
y2 + 10y + 25 = (y + 5)2
Half of +10 is +5, and that +5 is the constant inside (y + 5)2. Again — to keep the equation balanced, the 25 we added on the left has to appear on the right as well.
Step 4: balance both sides and combine
We added 9 and 25 to the left. We have to add the same numbers to the right:
(x2 − 6x + 9) + (y2 + 10y + 25) = −9 + 9 + 25
The right-hand side collapses to a single number:
−9 + 9 + 25 = 25
And the two groups on the left fold into the binomials we built:
(x − 3)2 + (y + 5)2 = 25
That is the standard form of the circle.
Step 5: read off the center and radius
Compare with the canonical standard form:
(x − h)2 + (y − k)2 = r2
Line them up term by term. The first parenthesis is (x − 3), which matches (x − h) when h = 3. The second is (y + 5), which we have to rewrite as (y − (−5)) to match (y − k) — that gives k = −5. The right-hand side is 25 = r2, so r = 5.
Center (3, −5) · Radius 5
The center sits in the fourth quadrant. The circle has radius 5, so it reaches from x = −2 to x = 8 horizontally and from y = −10 to y = 0 vertically. The y-axis itself (where x = 0) cuts through the circle; the x-axis grazes it tangent at the very top. Both observations make a quick sanity check: if you sketched this circle and it didn't touch the x-axis, something went wrong in the algebra.
The three mistakes worth memorizing
- Forgetting to take the square root of r2. Standard form gives you r2 = 25, not r = 25. Students who skip the square root write "radius 25" and lose the point. Pause and read the equation: it says r2.
- Sign-flipping the center. Standard form is (x − h)2 + (y − k)2. The numbers in the center are the values being subtracted inside the parentheses. (y + 5)2 means k = −5, not k = 5. Read every plus inside a squared binomial as "minus negative."
- Adding to one side only. Completing the square only works if both sides stay equal. If you add 9 and 25 to the left without adjusting the right, you've changed the equation. Every term added on the left has to land on the right too — that is where r2 comes from in the first place.
How SOMATH teaches completing the square
At SOMATH we drill completing the square as a single procedure — half, square, add to both sides — and we run it on three settings before the student is allowed to use it on a circle: (1) factoring a quadratic that doesn't factor cleanly, (2) finding the vertex of a parabola from y = ax2 + bx + c, and (3) deriving the quadratic formula from scratch. By the time a student gets to Stewart's Q3, the procedure is reflexive. That matters because in calculus you will see it again — every time a partial-fractions denominator has no real roots, every time an integrand needs to be rewritten as a shifted quadratic to match an arctangent or arcsine pattern, and every time you set up the equation of a conic from its general form.
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 do worksheet stacks. Each new technique gets introduced with the why, then practiced on three increasingly hostile problems, then connected to the calculus context where the student will meet it next. Stewart's Q3 takes one minute in class; the lesson around it takes thirty.
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FAQ
What is the standard form of a circle?
The standard form is (x − h)2 + (y − k)2 = r2, where (h, k) is the center and r is the radius. General form, by contrast, is x2 + y2 + Dx + Ey + F = 0. Completing the square is the bridge from general form to standard form.
Why do you take half of the linear coefficient and square it?
Because that is the constant that makes a quadratic into a perfect square. The expansion of (x + a)2 is x2 + 2ax + a2. The linear coefficient is 2a, so to recover a from a quadratic you've been handed, take half of the linear coefficient. The constant you add to complete the square is a2 — that is, the half, then squared.
How do you know if a general-form equation is actually a circle?
Complete the square. If the right-hand side comes out positive, it is a circle of radius sqrt of that number. If it comes out zero, the equation describes a single point (a "degenerate" circle). If it comes out negative, there is no real graph at all. Our Q3 lands at 25, which is positive, so we have a real circle of radius 5.
Does the order of x and y matter?
No. Addition is commutative, so it doesn't matter which group you complete first or which variable you write first inside standard form. We write x first by convention because that matches how (h, k) is read — h is the x-coordinate of the center, k is the y-coordinate.
Will this exact technique appear in AP Calculus?
Yes. Completing the square is the standard preprocessing step before partial fractions when a denominator has no real roots, before arctan-pattern integrals, and before sketching conics during transformations. The Stewart diagnostic is intentionally choosing it because the algebra in calculus assumes fluency here.
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