Question type

Circle equations on the SAT

A circle with centre $(h, k)$ and radius $r$ has equation $(x - h)^{2} + (y - k)^{2} = r^{2}$ . If the equation is expanded, complete the square in $x$ and in $y$ : $x^{2} + y^{2} - 10 x + 6 y - 2 = 0$ becomes $(x - 5)^{2} + (y + 3)^{2} = 36$ , so the radius is 6, not 36.

These questions give a circle's equation and ask for its centre or radius, give a centre and a point and ask for the equation, or ask whether a point lies on the circle.

It appeared on three of the five official practice tests we analysed.

On 3 of 5 official practice tests
20 practice questions of this type
Desmos helps

Why students miss it

Reading $r^{2}$ as $r$ . In $(x - 5)^{2} + (y + 3)^{2} = 36$ the radius is 6.

Sign of the centre: $(y + 3)^{2}$ means $k = - 3$ .

Completing the square on one side only: whatever you add to make the squares, add to the right-hand side too.

How to spot it

An equation with both $x^{2}$ and $y^{2}$ , with matching coefficients.
"Center", "radius", "lies on the circle", "tangent".
Endpoints of a diameter.

The method

Standard form

Group the $x$ terms and the $y$ terms, and move the constant to the right.

$x^{2} + b x$: add $(b /2)^{2}$ to both sides.
$y^{2} + d y$: add $(d /2)^{2}$ to both sides.
Right side: is $r^{2}$ : take its square root.
Diameter endpoints: centre is their midpoint, radius half their distance.

With Desmos

Desmos graphs the circle from any form of its equation.

Type the equation exactly as given, expanded form included: x^2+y^2-10x+6y-2=0.
Click the leftmost and rightmost points; the centre is halfway between them and the radius is half the distance.

Desmos does not label the centre; read it from the extreme points, or complete the square.

Two real questions

From the 1500prep bank, the way the app shows them. Answer first, then open the explanation.

1Question 1 of 2#73Geo

The equation x² + y² − 10x + 6y − 2 = 0 represents a circle in the xy-plane. What is the radius of the circle?

Pick an answer, then check it.

Show the answer and explanation

Complete the square on each variable.

x^{2} - 10 x = (x - 5)^{2} - 25

and

y^{2} + 6 y = (y + 3)^{2} - 9

. So

(x - 5)^{2} + (y + 3)^{2} - 34 - 2 = 0

, giving

(x - 5)^{2} + (y + 3)^{2} = 36

and

r = 36 = 6

2Question 2 of 2#45Geo

The circle in the xy-plane has equation (x − 7)² + (y + 2)² = 100. If the point (1, k) lies on the circle, what is the greatest possible value of k?

Pick an answer, then check it.

Show the answer and explanation

Plug

x = 1

in:

(1 - 7)^{2} + (k + 2)^{2} = 100

, so

36 + (k + 2)^{2} = 100

and

(k + 2)^{2} = 64

. Then

k + 2 = \pm 8

, giving

k = 6

k = - 10

. The greatest is

6

Questions students ask

How do you find the radius of a circle from its equation?

Write it as (x − h)² + (y − k)² = r², completing the square if needed, and take the square root of the right-hand side.

How do you complete the square for a circle equation?

For each variable, take half its linear coefficient, square it, and add it to both sides. x² − 10x becomes (x − 5)² after adding 25.

Practise this type until it is automatic

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