Find the general form of the equation of the circle.
Center at the point (-4,-3); containing the point (-3,3)
a) x^2+y^2+8x+6y-12=0
b) x^2+y^2+4x-6y+22=0
c) x^2+y^2-4x+6y+22=0
d) x^2+y^2+4x-6y+4=0
To find the general form of the equation of a circle, you need to know the coordinates of the center and a point on the circle. The general 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 of the circle is given as (-4, -3), and a point on the circle is given as (-3, 3).
1. Substitute the coordinates of the center into the equation:
(x - (-4))^2 + (y - (-3))^2 = r^2
(x + 4)^2 + (y + 3)^2 = r^2
2. Substitute the coordinates of the point (-3, 3) into the equation:
(-3 + 4)^2 + (3 + 3)^2 = r^2
1^2 + 6^2 = r^2
1 + 36 = r^2
37 = r^2
3. Now, substitute the value of r^2 in the equation:
(x + 4)^2 + (y + 3)^2 = 37
Comparing this equation with the given options:
a) x^2+y^2+8x+6y-12=0
b) x^2+y^2+4x-6y+22=0
c) x^2+y^2-4x+6y+22=0
d) x^2+y^2+4x-6y+4=0
The correct answer is option (c) x^2+y^2-4x+6y+22=0