Steply

Which conic is x^2 + y^2 - 6x - 7 = 0

Step by step, with the reason for every step · Conics from the general form · Precalculus

Step by step

The question

Question: name the conic and write in standard form x² + y² − 6x − 7 = 0.

x2 + y2 − 6x − 7 = 0

1 Name the conic

Look at the squares.

x² and y² both have 1 in front.

So it is a circle.

only x² or only y² → parabola
x² and y², same number in front → circle ✓
same signs, different numbers → ellipse
different signs → hyperbola

2 Group x and y

Put the x terms together in a bracket, and the y terms together.

(x2 − 6x) + y2 − 7 = 0

Move the plain number to the right: − 7 becomes + 7.

(x2 − 6x) + y2 = 7

3 Complete the square for x

Half of −6 is −3.

(−3) times (−3) is 9.

We add 9 inside the bracket.

So we add 9 on the right too.

(x2 − 6x + 9) + y2 = 7 + 9
−6 ÷ 2 = −3, (−3) × (−3) = 9

Careful!

What we add on the left, we add on the right too.

Then the two sides stay equal.

Careful: add 9 on both sides

4 Write as squares

Each bracket is now a full square.

On the right, 7 + 9 is 16.

(x − 3)2 + y2 = 16

It is the circle rule: center (3, 0), radius 4.

circle: center (3, 0), r = 4

5 Check

Let's check with a point on it: (7, 0), the right end of the circle.

Put it in the first equation.

7² + 0² − 6 × 7 − 7 = 49 + 0 − 42 − 7 = 0

We get 0, like the first equation says.

So the standard form is right!

Let's look back.

First: name the conic.

Then: group x and y.

Then: complete the square for x.

Then: write as squares.

Last: check.

Answer
(x − 3)2 + y2 = 16

Your turn

Try this one yourself, then check:

Name the conic and write in standard form x2 + y2 − 4x = 0
Show the answer
(x − 2)2 + y2 = 4

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