Circles in the Coordinate Plane

Circles in the Coordinate Plane | SATMath800
SAT Math • Coordinate Geometry

Circles in the Coordinate Plane

From the geometry behind a circle to the standard equation, general form, completing the square, and the fastest SAT shortcuts.

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r C(h,k) P(x,y)
01
START HERE

What does a circle mean?

A circle is the set of all points that are the same distance from one fixed point.

C

Center

The fixed point is called the center.

r

Radius

The constant distance from the center is the radius.

P

Every point

Every point on the circle is exactly r units from the center.

SAT connection: When the SAT gives you a circle in the coordinate plane, it is often testing whether you can connect this geometric definition to an equation.
02
COORDINATE SYSTEM

Let the center be \((h,k)\)

Instead of assuming that the center is at the origin, let the center be anywhere on the coordinate plane.

The center is represented by

\[ \boxed{(h,k)} \]

Here, \(h\) tells us the horizontal location and \(k\) tells us the vertical location.

Now choose any point \(P=(x,y)\) on the circle. The horizontal and vertical differences between \(P\) and the center are:

\[ x-h \qquad\text{and}\qquad y-k \]
03
THE BASIC PICTURE

A circle can be centered anywhere

Before memorizing an equation, look at what a circle actually means on the coordinate plane.

x y C(h, k) P(x, y) r x − h y − k
C

Center

The center can be anywhere on the coordinate plane: \((h,k)\).

r

Radius

Every point on the circle is exactly \(r\) units from the center.

P

A point on the circle

If \(P=(x,y)\), its horizontal and vertical distances from the center are \(x-h\) and \(y-k\).

The key idea: the circle equation is really just the Pythagorean theorem applied to the right triangle formed by the center and a point on the circle.
04
THE CONNECTION

From the triangle to the equation

The horizontal leg of the triangle has length \(x-h\), while the vertical leg has length \(y-k\). The radius is the hypotenuse.

\[ (x-h)^2+(y-k)^2=r^2 \]
Pythagorean theorem → equation of a circle
This is the standard form. It tells us the center and radius immediately.
05
STANDARD FORM

The equation to recognize instantly

Standard equation of a circle

\[ \boxed{(x-h)^2+(y-k)^2=r^2} \]

Center

Read directly from the equation: \[ (h,k) \]

Radius

Take the square root of the right side: \[ r=\sqrt{r^2} \]

Important sign rule

The signs inside the parentheses appear opposite to the center coordinates.

06
READ IT FAST

Extract the center and radius

Example 1 • Standard form
Find the center and radius of
\[ (x-3)^2+(y+4)^2=49 \]
Center: \[ (3,-4) \] Radius: \[ r=\sqrt{49}=7 \]
Watch the signs. \((x-3)^2\) means \(h=3\), while \((y+4)^2=(y-(-4))^2\) means \(k=-4\).
07
THE LONG WAY

Completing the square

Sometimes the SAT gives us a circle in general form instead of standard form.

The long way: convert the general form into standard form by completing the square for both the \(x\)-terms and the \(y\)-terms.
\[ x^2-10x+y^2+6y-15=0 \]
08
STEP 1

How do we create the squared form?

Look at the coefficient of the linear term. Divide it by \(2\), then square the result.

Work with the x-terms
\[ x^2-10x \]

The coefficient of \(x\) is \(-10\).

\[ \frac{-10}{2}=-5 \]

Therefore the squared form must be

\[ (x-5)^2 \]

because

\[ (x-5)^2=x^2-10x+25 \]
Repeat for the y-terms
\[ y^2+6y \]

The coefficient of \(y\) is \(6\).

\[ \frac{6}{2}=3 \]

Therefore:

\[ (y+3)^2 \]

because

\[ (y+3)^2=y^2+6y+9 \]
09
IMPORTANT

But where did those extra numbers come from?

Completing the square creates extra constants. We cannot simply add them to one side and ignore them.

\[ x^2-10x \quad\longrightarrow\quad (x-5)^2 \]

But

\[ (x-5)^2=x^2-10x+25 \]

So we have introduced \(+25\).

The rule: whatever we add to one side of an equation, we must also account for on the other side.
10
PUT IT TOGETHER

Complete the square correctly

\[ x^2-10x+y^2+6y-15=0 \]
1

Group the x and y terms

\[ (x^2-10x)+(y^2+6y)-15=0 \]
2

Add \(25\) and \(9\)

\[ (x^2-10x+25)+(y^2+6y+9) =15+25+9 \]
3

Rewrite each trinomial

\[ (x-5)^2+(y+3)^2=49 \]
We now have standard form.
\[ \boxed{(x-5)^2+(y+3)^2=49} \]
11
INTERPRET

Now the answer is visible

C

Center

\[ (5,-3) \]
r

Radius

\[ \sqrt{49}=7 \]

Standard form

\[ (x-5)^2+(y+3)^2=49 \]
12
SAT SHORTCUT

There is a faster way

If the equation has the form

\[ x^2+y^2+Dx+Ey+F=0 \]

you can find the center without fully completing the square.

The center shortcut

\[ \boxed{ \left(-\frac D2,-\frac E2\right) } \]

Take the coefficient of \(x\), divide by \(-2\). Then take the coefficient of \(y\), divide by \(-2\).

13
UNDERSTAND IT

Why does the shortcut work?

Start with the \(x\)-part of the general form:

\[ x^2+Dx \]

Completing the square tells us to divide \(D\) by \(2\), then square it.

\[ \left(\frac D2\right)^2 \]

The resulting squared expression is

\[ \left(x+\frac D2\right)^2 \]

Compare this with

\[ (x-h)^2 \]
Therefore, \[ h=-\frac D2. \] The exact same reasoning gives \[ k=-\frac E2. \]
14
EXAMPLE

Find the center quickly

SAT-style example
The equation of a circle is
\[ x^2+y^2-14x+8y-15=0 \]
What is the center?
Identify the coefficients:
\[ D=-14,\qquad E=8 \]
Apply the shortcut:
\[ \left( -\frac{-14}{2}, -\frac{8}{2} \right) \]
Therefore:
\[ \boxed{(7,-4)} \]
15
GO ONE STEP FURTHER

Finding the radius from general form

Once we know the coefficients \(D,E,F\), we can also determine the radius without writing every intermediate line.

\[ \boxed{ r^2= \frac{D^2+E^2}{4}-F } \]

Therefore,

\[ \boxed{ r= \sqrt{ \frac{D^2+E^2}{4}-F } } \]
Important: This shortcut assumes the \(x^2\) and \(y^2\) coefficients are both \(1\) and there is no \(xy\)-term.
16
EXAMPLE

Find the radius quickly

General-form radius
Consider
\[ x^2+y^2-14x+8y-15=0 \]
Here,
\[ D=-14,\quad E=8,\quad F=-15 \]
Therefore,
\[ r^2= \frac{(-14)^2+8^2}{4}-(-15) \]
\[ r^2= \frac{196+64}{4}+15 \]
\[ r^2=65+15=80 \]
So,
\[ \boxed{r=\sqrt{80}=4\sqrt5} \]
17
STRATEGY

Which method should you use?

1

Need the center?

Use \[ \left(-\frac D2,-\frac E2\right) \] when the equation is in general form.

2

Need the radius?

Use the radius formula if the problem only asks for the radius.

3

Need the whole equation?

Complete the square and convert the equation to standard form.

Computer Scientist’s SAT strategy

Do not perform ten algebraic steps when the question only requires one piece of information. Identify exactly what the SAT is asking for, then choose the shortest valid route.

18
WATCH OUT

Common circle mistakes

!

Flipping the signs

\((x-4)^2\) means \(h=4\), not \(-4\).

!

Forgetting the square root

If \[ r^2=49, \] then the radius is \(7\), not \(49\).

!

Ignoring added constants

Completing the square changes both sides. The extra values must be accounted for.

19
FINAL REVIEW

Your circle toolkit

Standard form

\[ \boxed{(x-h)^2+(y-k)^2=r^2} \]

Center from general form

\[ \boxed{ \left(-\frac D2,-\frac E2\right) } \]

Radius from general form

\[ \boxed{ r= \sqrt{ \frac{D^2+E^2}{4}-F } } \]

Remember the structure

Center → distances → Pythagorean theorem → circle equation.

Once you understand that chain, the formulas stop being isolated facts and become different ways of describing the same geometric idea.

20
SAT READY

See the circle. Then choose the shortest path.

A circle in the coordinate plane is not just an equation. It is a geometric relationship between a center, a radius, and every point on the circle.

\[ \boxed{ (x-h)^2+(y-k)^2=r^2 } \]

When the SAT gives you general form, recognize the coefficients and decide whether the long way or the shortcut is more efficient.

One final habit

Don’t solve the equation just because you can. Solve exactly what the question asks.

SAT Circle Questions

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