Chapter 4 · Geometry and Equations · Concept 3
4-19 Conditions for Forming a Circle
An equation that looks like a circle is not always a circle. Complete the square, and whatever number lands on the right has to be positive — because that number is the radius squared, and nothing real has a negative square.
Guiding question
When does a circle stop being a circle?
Learning outcome. Identify the conditions required for an equation to represent a circle.
One test, applied at the very end
Turn x² + y² + lx + my + n = 0 into
(x − a)2 + (y − b)2 = constant
it is a circle exactly when constant > 0
And what the other two cases are.
= 0 — the radius is zero, so only the centre itself fits. One lonely point, not a circle.
< 0 — two squares can never add up to a negative, so no point anywhere works. The equation describes nothing at all.
This is why the rule says strictly greater than zero. Reading it as “≥” lets the single point in, and it loses you the boundary of every answer.
The answer is a range, so it is an inequality.
The constant usually comes out as a quadratic in k, and you need it positive — so this is the same job as Chapter 1. A quadratic that opens upward is positive outside its two roots, which gives an answer in two pieces: k < −1 or k > 3. Writing it the other way round, as −1 < k < 3, is the commonest wrong answer here.
This is the Warm Up equation. Slide k from the right and watch the circle shrink, become a single dot, and then disappear entirely.
x² + y² − 2kx + 2ky + 4k + 6 = 0
(x − k)² + (y + k)² = 2k² − 4k − 6
k = 4.0
right-hand side = 2k² − 4k − 6 = 10.00
Positive, so this is a real circle with radius √10.00 ≈ 3.16.
The two collapse points are at k = −1 and k = 3. Between them there is no circle.
Find the values of k for which x² + y² − 2kx + 2ky + 4k + 6 = 0 is a circle.
- 1
Complete the square, treating k as an ordinary number.
(x − k)2 − k2 + (y + k)2 − k2 + 4k + 6 = 0Half of −2k is −k, and half of +2k is +k. Each bracket leaves a k² behind, so two k² come off.
Test one number from each side.
At k = 0 (inside the gap): 2(0) − 0 − 6 = −6, negative — no circle, correct. At k = 4 (outside): 32 − 16 − 6 = 10, positive — a circle, correct. Two substitutions, and the direction of the inequality is settled beyond doubt.
Work with a partner. After you agree on an answer, each pick a number — one from inside your range and one from outside — and substitute. If the inside one gives a negative constant, your inequality is facing the wrong way.
An automated ultrasonic welder generates an acoustic stress boundary modelled by x² + y² + 6x − 2ky + 2k² − 4k + 9 = 0. For a stable safety perimeter to exist around the machine, the radius must be a real, positive number. Find the exact range of the power setting k that guarantees a real circle.
Facing the inequality the wrong way.
An upward parabola is positive OUTSIDE its roots and negative BETWEEN them. Sketch it, or test one number from each region — do not guess.
Forgetting that dividing by a negative flips it.
Exercise 3 turns −2k² + 6k + 8 > 0 into k² − 3k − 4 < 0. Miss the flip and you get the exact opposite range.
Allowing the constant to be zero.
A zero right-hand side gives a single point, not a circle. The inequality is strictly greater than, so the boundary values are excluded.
Dropping a leftover when k is inside the square.
Completing the square on −2kx leaves −k² behind, and that k² must be added to the right along with everything else.
Solving an equation instead of an inequality.
The question asks for a RANGE. Finding the two roots is only halfway — you still have to say which side of them the answer lives on.
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