Four curves come from one cone.
Tilt the cutting plane far enough and it meets both halves of the cone at once, giving the two branches of a hyperbola. Circle, ellipse, parabola, hyperbola: one family, four angles.
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The Circle
Every conic is a slice through a cone. The circle is the straight cut.
Identify the graph
x² + y² − 6x + 4y + 13 = 0
Complete the square twice: (x − 3)² − 9 + (y + 2)² − 4 + 13 = 0, so (x − 3)² + (y + 2)² = 0. Only x = 3 and y = −2 satisfy that, so the graph is the single point (3, −2) — a circle of radius zero. A and B read the constant as r² without completing the square first, which is the step that decides everything here.
The standard form is the distance formula in disguise.
(x − h)² + (y − k)² = r² says "every point is r away from (h, k)".
(x − 3)² + (y + 2)² = 25
centre (3, −2)
radius 5, not 25
x² + y² = 49
centre (0, 0)
radius 7
(x + 1)² + y² = 3
centre (−1, 0)
radius √3
Two sign flips and one square root
The signs inside the brackets reverse when you read off the centre, and the number on the right is r², not r. Three chances to slip in a single line, and all three are worth checking before moving on.
General form becomes standard by completing the square twice.
x² + 10x + y² − 8y = 8
group and move the constant
(x² + 10x + 25) + (y² − 8y + 16) = 8 + 25 + 16
add to BOTH sides
(x + 5)² + (y − 4)² = 49
standard form at last
centre (−5, 4), radius 7
read it off
Halve the coefficient of x and square it; do the same for y. Whatever you add on the left must be added on the right, or the equation stops being the same equation.
The right-hand side decides whether there is a circle at all.
= 25
a genuine circle
radius 5
= 0
a single point
the centre only
= −4
no graph
two squares cannot total a negative
These last two are the degenerate cases. They look like circles all the way through the algebra and only reveal themselves at the final line, which is why the answer "a point" or "no graph" is a real answer and not a mistake.
Variations: a leading 2, halves, and a point on the circle.
The same completing-the-square, in the three places it most often stalls.
A 2 in front of x² and y²: 2x² + 2y² − 8x + 12y − 6 = 0
x² + y² − 4x + 6y − 3 = 0
divide EVERY term by 2 first
(x² − 4x + 4) + (y² + 6y + 9) = 3 + 4 + 9
complete both squares
(x − 2)² + (y + 3)² = 16
standard form
centre (2, −3), radius 4
√16
Completing the square only works when x² and y² have coefficient 1, so divide first. Equal coefficients on x² and y² are what make it a circle at all.
Odd coefficients, so halves appear: x² + y² − 3x + y − 32 = 0
(x² − 3x + 94) + (y² + y + 14) = 32 + 94 + 14
half of −3 is −32; half of 1 is 12
(x − 32)² + (y + 12)² = 164 = 4
64 + 94 + 14
centre (32, −12), radius 2
keep the halves as fractions
A fractional centre is normal. Rounding 32 to 2 would give a wrong circle.
The centre and one point are given: centre (−1, 4), passing through (2, 0)
r² = (2 + 1)² + (0 − 4)² = 9 + 16 = 25
distance from centre to the point, squared
(x + 1)² + (y − 4)² = 25
no need to take the root at all
Here the work runs backwards: we build the equation instead of reading it. The squared distance is r² directly.
A diameter gives the centre and the radius at once.
centre = midpoint = (3, 4)
average the coordinates
diameter = √36 + 64 = 10
distance between the ends
radius = 5, so r² = 25
halve it BEFORE squaring
(x − 3)² + (y − 4)² = 25
the equation
On the exam
Tests give the general form and ask for the centre and radius, or give a diameter or a centre and a point and ask for the equation; entrance tests like the degenerate "a point" or "no graph" choices. The fastest route is divide, complete both squares, then look at the right-hand side before taking any root. A 10-second check: substitute the top point (h, k + r) into the original equation; it must give 0.
Now you try
Practice problem
Find the centre and radius
x² + y² + 8x − 2y + 8 = 0
(x² + 8x + 16) + (y² − 2y + 1) = −8 + 16 + 1, so (x + 4)² + (y − 1)² = 9. That gives centre (−4, 1) and radius 3. A forgets to flip the signs when reading the centre. C reports r² as the radius.
Hint: move the 8 across first, then add 16 and 1 to both sides. The right-hand side ends up as −8 + 17.