The distance formula, squared and rearranged — and the one move that turns the hard version into the easy one.
What is the value of the radius?
In the xy-plane, the graph of
x² + y² − 10x + 6y = 2
is a circle. What is the radius of the circle?
Complete the square on both variables: (x² − 10x + 25) + (y² + 6y + 9) = 2 + 25 + 9, so (x − 5)² + (y + 3)² = 36 and the radius is √36 = 6. A adds 25 and 9 only on the left, leaving r² = 2 — the single most common slip here. C reports the diameter.D stops at r² and never takes the square root.
1
A circle is every point that sits the same distance from one centre.
Say that with the distance formula and the equation writes itself.
√(x − h)² + (y − k)² = r
distance from (x, y) to the centre (h, k)
(x − h)² + (y − k)² = r²
square both sides — the radical is gone
(h, k)
the centre
read it out of the brackets
r²
the radius, squared
the right-hand side, never the radius itself
This is the standard form, sometimes called centre-radius form. Every circle
question on the SAT is about getting an equation into it, or reading one that is
already there.
2
Read the centre out of the brackets — with the sign flipped.
The form subtracts h and k, so whatever you see inside a bracket is the opposite of the coordinate.
(x + 4)² is (x − (−4))²
so h = −4
(y − 7)² is already (y − (7))²
so k = 7
r² = 100, so r = 10
the right side is the square, not the radius
Equation
Centre
Radius
(x + 4)² + (y − 7)² = 100
(−4, 7)
10
(x − 3)² + (y + 2)² = 25
(3, −2)
5
x² + y² = 81
(0, 0)
9
(x − 1)² + (y + 5)² = 12
(1, −5)
2√3
The bracket shows the opposite sign
(x + 4)² gives h = −4, not 4. Reading a centre of (4, −7) off the
first row is the standard error in this topic, and the SAT always puts that
sign-flipped centre in the choices. A bare x², with no bracket at all, means h = 0.
The last row is a reminder that r² does not have to be a perfect square.
r = √12 = 2√3 ≈ 3.46 is a perfectly good radius.
(0, 0)9 + 4 = 13 < 25inside — the origin is not on this circle
Substituting a point and comparing with r² is the fastest test of whether that
point is on the circle, inside it, or outside it. Bigger than r² means outside.
4
Going the other way: a centre and a radius give you the equation.
A circle in the xy-plane has centre (−6, 4) and radius 9. Which equation represents the circle?
h = −6, k = 4, r = 9
given
(x − (−6))² + (y − 4)² = 9²
straight substitution, brackets and all
(x + 6)² + (y − 4)² = 81
minus a minus, and 9² = 81
(x + 6)² + (y − 4)² = 81
Square the radius, then stop
Writing = 9 instead of = 81 describes a circle of radius 3. Writing
(x − 6)² instead of (x + 6)² moves the centre twelve units across. Those two
slips are exactly the two wrong answers the SAT offers beside the right one.
5
Expanded equation? Complete the square twice.
This is the hard version, and it is the one from the opening problem: x² + y² − 10x + 6y = 2.
(x² − 10x) + (y² + 6y) = 2
group the x's, group the y's, constants on the right
half of −10 is −5, and (−5)² = 25
the number that finishes the x bracket
half of 6 is 3, and 3² = 9
the number that finishes the y bracket
(x² − 10x + 25) + (y² + 6y + 9) = 2 + 25 + 9
add both to both sides
(x − 5)² + (y + 3)² = 36
fold each bracket; use the half, not the whole
centre (5, −3), radius 6
√36 = 6
radius = 6
x term
−10x
half it
−5
square it
25
Both sides, both times
The 25 and the 9 have to be added on the right as well. Adding them only on the
left leaves r² = 2 and a radius of about 1.41 — a completely different circle, and
choice A on the opening problem.
Desmos will draw it for you
The Digital SAT gives you the graphing calculator the whole time. Type
x^2 + y^2 − 10x + 6y = 2 exactly as written and Desmos plots the circle; click
the curve to read points off the rim, or check your centre by typing the point
(5, −3) and seeing it land in the middle. Complete the square by hand when the
numbers are clean, and let the calculator confirm.
6
A point on the circle hands you the radius.
A circle in the xy-plane has its centre at (2, 5) and passes through the point (7, 17). Which equation represents the circle?
r = √(7 − 2)² + (17 − 5)²
the radius is the distance from centre to rim
r = √25 + 144 = √169 = 13
5 and 12 — a familiar right triangle
(x − 2)² + (y − 5)² = 169
r² = 13² = 169
Faster still: you never needed r on its own. Substitute the point straight into
(7 − 2)² + (17 − 5)² and the answer, 169, is r². Squaring and un-squaring
is wasted work.
The same idea covers the other common version: given the two endpoints of a
diameter, the centre is their midpoint and the radius is half the distance
between them. Take the whole distance as the radius and the circle comes out twice
too wide.
Group · Halve and square · Add to both sides · Root the right side
The brackets always show the opposite of the centre's coordinates.
Now you try
Practice problem
What is the value of the diameter?
In the xy-plane, the graph of
x² + y² − 4x + 16y = −4
is a circle. What is the length of a diameter of the circle?
Half of −4 is −2 and (−2)² = 4; half of 16 is 8 and 8² = 64. So (x² − 4x + 4) + (y² + 16y + 64) = −4 + 4 + 64, giving (x − 2)² + (y + 8)² = 64. Then r = 8 and the diameter is 2r = 16. A answers the radius, the quantity the question did not ask for. C stops at r² = 64.D doubles r² instead of r.
Hint: the constant on the right starts out negative, and that is fine — what
matters is the total after you add both completing-square numbers. Get to standard
form first, then read off r, then check which length the question actually wanted.
