In the coordinate plane, a circle centered on point (-3, -4) passes th

The circle is with center (-3,-4) and one of the point as (1,1).Using distance equation we can get the radius :
r^2 = sqrt ((-3-1)^2 + (-4-1)^2) or r = 41^1/2

Hence aread = pi (r)^2 = 41 pi

From the distance formula:

r^2= (-3-1)^2 + (-4-1)^2
r^2= 16+25=41

As we know, Area of a Circle=pi * r^2
Therefore, Area of Circle= 41 pi
Hence, the answer is 'E'

VERITAS PREP OFFICIAL SOLUTION:

As you know, the area of a circle is \(π(r)^2\), meaning that your goal is to find the radius of this circle. To go from the center of the circle (-3, -4) to the given point on the circle (1, 1), you move 4 spaces horizontally (from x-coordinate -3 to x-coordinate 1) and 5 spaces vertically (from y-coordinate -4 to y-coordinate 1). That sets up a right triangle in which the hypotenuse is the radius. a^2+b^2=c^2 becomes 4^2+5^2=c^2, so c^2=16+25=41. And here's where a shortcut awaits. Since your job is to take the radius (c in the Pythagorean Theorem) and square it, then multiply by π (π(r)^2) you're actually about done. Since \(r^2=41\), the answer is \(41π\).

Apply distance formula between the xy pair of (-3-,4) and (1,1)
The distance so obtained will be the radius of the circle
Once radius is known then \(area = π*r^2\)

\(D= \sqrt{(-3-1)^2+(-4-1)^2}\)
\(D= \sqrt{(-4)^2+(-5)^2}\)

\(D= \sqrt{16+25}\)

\(D= \sqrt{41}\)

\(Radius = \sqrt{41}\)
\(Radius^2 = 41\)

\(Area = Radius^2*π\)

\(Area= 41π\)

Answer is E