The equation of the circle with center 2,-3 and radius 4 is
Question 7 Select one:
y2+x2-4x-6y-3=0
y2+x2-4x+6y-3=0
y2-x2-4x+6y-3=0
y2+x2+4x+6y-3=0
The equation of a circle with center (h, k) and radius r is (x-h)^2 + (y-k)^2 = r^2.
In this case, the center is (2, -3) and the radius is 4.
Plugging these values into the equation of a circle, we get:
(x-2)^2 + (y-(-3))^2 = 4^2
(x-2)^2 + (y+3)^2 = 16
Expanding the equation, we get:
x^2 - 4x + 4 + y^2 + 6y + 9 = 16
x^2 + y^2 - 4x + 6y + 13 = 16
Rearranging the terms, we get:
x^2 + y^2 - 4x + 6y - 3 = 0
Therefore, the correct equation of the circle is y^2 + x^2 - 4x + 6y - 3 = 0.
So, the answer is:
y^2 + x^2 - 4x + 6y - 3 = 0