what is the equation of the circle with center -4,-3 passes through the point 6,2?The answer choices are A(x-4)^2+(y-3)^2=25,B(x-4)^2+(y-3)^2=125,C(x-(-4))^2+(y-(-3))^2=25 or D(x-(-4))^2+(y-(-3))^2=125.

Question

mathmate · Answer

Hint:
Equation of a circle with radius r and  centre O(x0,y0) is
$(x-x0)^2+(y-y0)^2=r^2$
Example:
A circle with radius 6 with centre O(-2,3) has the equation
$(x-(-2))^2+(y-3)^2=6^2$    or
$(x+2)^2+(y-3)^2=6^2$

mathmate · Answer

Additional hint: The radius is the distance between the centre O(-4,-4) and the given point P(6,2) To find distance between two points, try: http://www.purplemath.com/modules/distform.htm

campbell_st · Answer

the general equation of a circle is 

$$(x - h)^2 + (y - k)^2 = r^2$$

the centre is (h, k) anf r = radius

you know the centre is (-4, -3)

so the equation becomes

$$(x + 4)^2 + (y + 3)^2 = r^2$$

to find the value of r^2, subsitute the point the circle passes through into the equation, the point is (6, 2) 

so 

$$(6 + 4)^2 + (2 + 3)^2 = r^2$$

now solve for r^2

hope it helps