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.

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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.

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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 \((x+2)^2+(y-3)^2=6^2\)
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
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

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