A circle is every point a fixed distance from the centre — which is the distance formula, squared.
Read the equation
What are the centre and radius of (x − 3)² + (y + 5)² = 49?
In (x − h)² + (y − k)² = r², the centre is (h, k) and the radius is √r². The brackets read (x − 3) and (y + 5), so h = 3 and k = −5, and r = √49 = 7. A leaves the radius as 49, which is r². D flips both signs — the rule flips each sign once, so (x − 3) gives +3 and (y + 5) gives −5.
(x − h)² + (y − k)² = r²
The dashed legs are the horizontal and vertical gaps, x − h and y − k. The radius is the hypotenuse, so Pythagoras gives the equation directly. It is the distance formula with both sides squared.
| Equation | Centre | Radius |
|---|---|---|
| (x − 3)² + (y + 5)² = 49 | (3, −5) | 7 |
| (x + 2)² + (y − 1)² = 20 | (−2, 1) | 2√5 |
| x² + y² = 16 | (0, 0) | 4 |
Both traps live in the same line. The bracket's sign reverses to give the centre, and the number on the right is r², so it must be rooted to give the radius.
centre (−4, 2), radius 6
substitute
(x + 4)² + (y − 2)² = 36
signs flip, radius squares
centre (1, 3), through (9, 9)
radius unknown
r² = (9 − 1)² + (9 − 3)² = 100
the distance, squared
(x − 1)² + (y − 3)² = 100
r = 10, but r² is what goes in
You never need the radius itself here — r² is what the equation wants, so stopping before the square root saves a step.
x² + y² + 10x − 2y − 10 = 0
general form
(x² + 10x) + (y² − 2y) = 10
group and move the constant
(x² + 10x + 25) + (y² − 2y + 1) = 10 + 25 + 1
add to BOTH sides
(x + 5)² + (y − 1)² = 36
centre (−5, 1), radius 6
Two separate completions here — 25 for the x group and 1 for the y group — and both must appear on the right as well, or the radius comes out wrong.
Is (−1, 4) on (x + 5)² + (y − 1)² = 36?
test it
16 + 9 = 25 < 36
inside the circle
Centre = midpoint of the diameter; radius = half its length.
A diameter has endpoints (−1, 1) and (3, 5). Write the equation of the circle.
centre (−1 + 32, 1 + 52) = (1, 3)
average the coordinates
diameter² = 4² + 4² = 32
gaps of 4 and 4
r² = 32 ÷ 4 = 8
r = d2, so r² = d²4
(x − 1)² + (y − 3)² = 8
check (3, 5): 4 + 4 = 8 ✓
Using the full distance as the radius gives (x − 1)² + (y − 3)² = 32, a circle twice as wide. The quick test catches it: an endpoint must satisfy the equation, and (3, 5) gives 8, not 32.
The same two forms; each case changes one detail.
Centre on the x-axis: centre (−6, 0), radius 2√3
(x + 6)² + (y − 0)² = (2√3)²
k = 0
(x + 6)² + y² = 12
(2√3)² = 4 × 3
Different: a zero coordinate leaves a bare y², and an irrational radius still squares to a whole number.
A coefficient on x² and y²: 2x² + 2y² + 4x − 12y − 12 = 0
x² + y² + 2x − 6y − 6 = 0
divide every term by 2 first
(x² + 2x + 1) + (y² − 6y + 9) = 6 + 1 + 9
complete both squares
(x + 1)² + (y − 3)² = 16
centre (−1, 3), radius 4
Different: x² and y² must have coefficient 1 before the halves are taken.
Not a circle at all: x² + y² − 4x + 2y + 9 = 0
(x² − 4x + 4) + (y² + 2y + 1) = −9 + 4 + 1
complete both squares
(x − 2)² + (y + 1)² = −4
no circle
Different: r² cannot be negative, so no point fits. If the right side were 0, the "circle" would shrink to the single point (2, −1).
Expect to read the centre and radius, to write the equation from a centre and a radius or a point, and to convert a general form by completing the square. The fastest route is to flip the bracket signs for the centre and take √ of the right side for the radius. The 10-second check: substitute a point we know is on the circle — it must give r² exactly.
Write the equation
A circle has centre (2, −6) and radius 3. What is its equation?
The centre's signs flip going into the brackets: h = 2 gives (x − 2) and k = −6 gives (y + 6). The right side is r² = 3² = 9. C leaves the radius unsquared and B flips the signs the wrong way. Check by substituting the centre's own coordinates: 0 + 0 = 0, which is less than 9, so the centre is inside its circle ✓
Hint: two things change going into the equation — each coordinate's sign flips, and the radius gets squared.
