Answer:
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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
