The distance formula, squared and rearranged — nothing more.
Write the equation
Write the equation of the circle with centre (3, −2) and radius 5.
The form is (x − h)² + (y − k)² = r². With k = −2, y − (−2) becomes y + 2, and the right side is r² = 25, not 5. C forgets to square the radius; B flips both signs.
√(x − h)² + (y − k)² = r
the distance formula
(x − h)² + (y − k)² = r²
square both sides
the centre
the radius, squared
That is the centre-radius form, and it hands you both facts at a glance.
h = 3, k = −2, r = 5
given
(x − 3)² + (y − (−2))² = 5²
straight substitution
(x − 3)² + (y + 2)² = 25
minus a minus
(y + 2)² means k = −2. Reading a centre of (3, 2) off that equation is the standard error in this topic.
And the right-hand side is r², not r. A 25 means a radius of 5.
Centre and a point on the rim. The circle has centre (−1, 3) and passes through (5, 11).
r² = (5 − (−1))² + (11 − 3)² = 36 + 64 = 100
the radius is the distance from centre to that point
(x + 1)² + (y − 3)² = 100
keep r² — no need to take the root
x² − 6x + y² + 4y = 12
group x's and y's, move the number
x² − 6x + 9 + y² + 4y + 4 = 12 + 9 + 4
halve and square each
(x − 3)² + (y + 2)² = 25
fold both brackets
centre
(3, −2)
radius
5
The 9 and the 4 must also be added on the right. Adding them only on the left gives r² = 12 and a radius of 3.46 — a different circle entirely.
| Equation | Centre | Radius |
|---|---|---|
| (x − 3)² + (y + 2)² = 25 | (3, −2) | 5 |
| x² + y² = 49 | (0, 0) | 7 |
| (x + 1)² + (y − 4)² = 12 | (−1, 4) | 2√3 |
A(1, 2) and B(7, 10) are opposite ends. Find the circle.
centre = midpoint = (4, 6)
average both coordinates
diameter = √6² + 8² = 10
the distance formula
radius = 5
half the diameter
(x − 4)² + (y − 6)² = 25
the equation
The distance between the two points is the diameter. Using 10 as the radius makes a circle four times too big in area.
"Touches" (is tangent to) a line means the radius is the straight distance from the centre to that line. For the axes and for lines x = c or y = c, that is one subtraction.
| Centre (−4, 3) touches | Radius | Equation |
|---|---|---|
| the x-axis | 3 (the y-coordinate) | (x + 4)² + (y − 3)² = 9 |
| the y-axis | 4 (the size of the x-coordinate) | (x + 4)² + (y − 3)² = 16 |
| the line y = −2 | 3 − (−2) = 5 | (x + 4)² + (y − 3)² = 25 |
The distance down to the x-axis is how high the centre sits — its y-value. Using the x-value is the trap answer every time.
Backwards: for what k is x² + y² + 8x − 4y + k = 0 a circle of radius 5?
(x² + 8x + 16) + (y² − 4y + 4) = −k + 16 + 4
complete both squares; k moves right
(x + 4)² + (y − 2)² = 20 − k
the right side is r²
20 − k = 25, k = −5
set it equal to 5², not 5
A leading 2. Find the centre and radius of 2x² + 2y² − 8x + 12y − 6 = 0.
x² + y² − 4x + 6y − 3 = 0
÷ 2 first: x² and y² must stand alone
(x² − 4x + 4) + (y² + 6y + 9) = 3 + 4 + 9
complete both squares; add on both sides
(x − 2)² + (y + 3)² = 16
fold both brackets
centre (2, −3), radius 4
the 16 is r²
What is different: completing the square needs 1x², so divide every term by the 2 before anything else.
An odd coefficient. Find the centre and radius of x² + y² + 5x − 6y + 3 = 0.
(x² + 5x + 254) + (y² − 6y + 9) = −3 + 254 + 9
half of 5 is 52, and (52)² = 254
(x + 52)² + (y − 3)² = 494
−3 + 9 = 6 = 244, and 244 + 254 = 494
centre (−52, 3), radius 72
root the top and the bottom
What is different: an odd x-coefficient puts a fraction in the centre. Keep it in quarters — rounding 254 to 6 changes the circle.
When it is not a circle. Is x² + y² − 4x + 2y + 9 = 0 a circle?
(x² − 4x + 4) + (y² + 2y + 1) = −9 + 4 + 1
the same two squares
(x − 2)² + (y + 1)² = −4
r² would be negative
no points at all
two squares can never add to a negative
What is different: the right side decides. Positive: a circle. Zero (a constant of 5 instead of 9): only the single point (2, −1). Negative: no graph.
Taking √4 and answering "radius 2" ignores the sign. A circle needs r² > 0; a negative right side means no circle exists.
Worked: the line y = x − 1 cuts the circle x² + y² = 13 at P and Q. Find PQ.
x² + (x − 1)² = 13
substitute the line into the circle
2x² − 2x − 12 = 0 → x² − x − 6 = 0
expand, collect, ÷ 2
x = 3 or x = −2 → P(3, 2), Q(−2, −3)
y from the line, y = x − 1
PQ = √5² + 5² = 5√2
the distance formula
Worked: a circle passes through (2, 1), (8, 1) and (2, 9). Find it.
(2, 1)→(8, 1) is horizontal, (2, 1)→(2, 9) is vertical
a right angle at (2, 1)
diameter: (8, 1) to (2, 9)
a right angle in a circle stands on a diameter
centre (5, 5), diameter √36 + 64 = 10, r = 5
midpoint, then half the length
(x − 5)² + (y − 5)² = 25
check (2, 1): 9 + 16 = 25 ✓
Circle items test reading the centre and radius (often from general form), and then one twist: a tangent line giving r, a missing constant, a chord, or three points with a right angle. The ten-second check is to put a point you know is on the circle into your final equation — it must give exactly r².
Write the equation
Write the equation of the circle with centre (−4, 1) and radius 6.
h = −4 gives x − (−4) = x + 4, and k = 1 gives y − 1. Then r² = 36. The signs inside the brackets are always the opposite of the centre's coordinates — that flip is the whole trick.
