Question

Determine the graph of the equation 16^2 + 16^2 − 16 + 24 − 61 = 0.

Answer 1

Answer:

Enter a problem...

Precalculus Examples

Popular Problems Precalculus Find the Properties 16x^2+16y^2-16x+24y-3=0

16

x

2

+

16

y

2

−

16

x

+

24

y

−

3

=

0

Add

3

to both sides of the equation.

16

x

2

+

16

y

2

−

16

x

+

24

y

=

3

Divide both sides of the equation by

16

.

x

2

+

y

2

−

x

+

3

y

2

=

3

16

Complete the square for

x

2

−

x

.

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(

x

−

1

2

)

2

−

1

4

Substitute

(

x

−

1

2

)

2

−

1

4

for

x

2

−

x

in the equation

x

2

+

y

2

−

x

+

3

y

2

=

3

16

.

(

x

−

1

2

)

2

−

1

4

+

y

2

+

3

y

2

=

3

16

Move

−

1

4

to the right side of the equation by adding

1

4

to both sides.

(

x

−

1

2

)

2

+

y

2

+

3

y

2

=

3

16

+

1

4

Complete the square for

y

2

+

3

y

2

.

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(

y

+

3

4

)

2

−

9

16

Substitute

(

y

+

3

4

)

2

−

9

16

for

y

2

+

3

y

2

in the equation

x

2

+

y

2

−

x

+

3

y

2

=

3

16

.

(

x

−

1

2

)

2

+

(

y

+

3

4

)

2

−

9

16

=

3

16

+

1

4

Move

−

9

16

to the right side of the equation by adding

9

16

to both sides.

(

x

−

1

2

)

2

+

(

y

+

3

4

)

2

=

3

16

+

1

4

+

9

16

Simplify

3

16

+

1

4

+

9

16

.

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(

x

−

1

2

)

2

+

(

y

+

3

4

)

2

=

1

This is the form of a circle. Use this form to determine the center and radius of the circle.

(

x

−

h

)

2

+

(

y

−

k

)

2

=

r

2

Match the values in this circle to those of the standard form. The variable

r

represents the radius of the circle,

h

represents the x-offset from the origin, and

k

represents the y-offset from origin.

r

=

1

h

=

1

2

k

=

−

3

4

The center of the circle is found at

(

h

,

k

)

.

Center:

(

1

2

,

−

3

4

)

These values represent the important values for graphing and analyzing a circle.

Center:

(

1

2

,

−

3

4

)

Radius:

1

image of graph