the point (4,7) lies on the circle centered at (-3,-2). find the equation of the circle in standard form.?

Answer 1

radius of the circle is distance betwwen the center and the point
r=sqrt(7^2+9^2)
r^2=49+81
r^2=130
the equation of the circle in standard form is
(x - h)^2 + (y - k)^2 = r^2
where (h,k) is the center of the circle
hence the equation is
(x +3)^2 + (y +3)^2 = 130

Answer 2

r ² = (4 + 3) ² + (7 + 2) ²
r ² = 49 + 81
r ² = 130
(x + 3) ² + (y + 2) ² = 130

If desired, may be written as:-
x ² + 6 x + 9 + y ² + 4 y + 4 = 130
x ² + y ² + 6 x + 4 y - 117 = 0

Answer 3

In general form the equation of a cirlce is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) = centre
radius = sq rt{[4 - -3]^2 + [7 - -2]^2}
r^2 = 7^2 + 9^2 = 130

so equation is
( x + 3)^2 + (y+2 )^2 = 130

Answer 4

(4+3)^2+(7+2)^2=r^2