The graph of
\[r = -2 \cos \theta + 6 \sin \theta\]is a circle.  Find the area of the circle.
From the equation $r = -2 \cos \theta + 6 \sin \theta,$

\[r^2 = -2r \cos \theta + 6r \sin \theta.\]Then $x^2 + y^2 = -2x + 6y.$  Completing the square in $x$ and $y,$ we get
\[(x + 1)^2 + (y - 3)^2 = 10.\]Thus, the graph is the circle centered at $(-1,3)$ with radius $\sqrt{10}.$  Its area is $\boxed{10 \pi}.$

[asy]
unitsize(0.5 cm);

pair moo (real t) {
  real r =-2*cos(t) + 6*sin(t);
  return (r*cos(t), r*sin(t));
}

path foo = moo(0);
real t;

for (t = 0; t <= pi + 0.1; t = t + 0.1) {
  foo = foo--moo(t);
}

draw(foo,red);

draw((-5,0)--(3,0));
draw((0,-1)--(0,7));

label("$r = -2 \cos \theta + 6 \sin \theta$", (6,5), red);
[/asy]
10\pi
Level 3
Precalculus
Problem: The graph of
\[r = -2 \cos \theta + 6 \sin \theta\]is a circle.  Find the area of the circle.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the area of the circle represented by the polar equation \( r = -2 \cos \theta + 6 \sin \theta \), we first need to convert this polar equation into Cartesian coordinates. The conversion formulas are \( x = r \cos \theta \) and \( y = r \sin \theta \). 

We can express \( r \) in terms of \( x \) and \( y \) using the relationships \( r = \sqrt{x^2 + y^2} \), \( \cos \theta = \frac{x}{r} \), and \( \sin \theta = \frac{y}{r} \). Substituting these into the equation gives:

\[
r = -2 \left(\frac{x}{r}\right) + 6 \left(\frac{y}{r}\right)
\]

Multiplying through by \( r \) to eliminate the denominator results in:

\[
r^2 = -2x + 6y
\]

Substituting \( r^2 = x^2 + y^2 \) leads to:

\[
x^2 + y^2 = -2x + 6y
\]

Rearranging this equation yields:

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

Next, we complete the square for both \( x \) and \( y \):

For \( x \):
\[
x^2 + 2x = (x + 1)^2 - 1
\]

For \( y \):
\[
y^2 - 6y = (y - 3)^2 - 9
\]

Substituting these completed squares back into the equation gives:

\[
(x + 1)^2 - 1 + (y - 3)^2 - 9 = 0
\]

Simplifying this results in:

\[
(x + 1)^2 + (y - 3)^2 = 10
\]

This is the standard form of a circle with center at \((-1, 3)\) and radius \( \sqrt{10} \).

The area \( A \) of a circle is calculated using the formula:

\[
A = \pi r^2
\]

Substituting the radius:

\[
A = \pi (\sqrt{10})^2 = 10\pi
\]

Thus, the final answer is \( 10\pi \).<|END_OF_LATENT|> The final answer is $10\pi$.