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billyjean
 one year ago
what is the equation of the circle with center 4,3 passes through the point 6,2?The answer choices are A(x4)^2+(y3)^2=25,B(x4)^2+(y3)^2=125,C(x(4))^2+(y(3))^2=25 or D(x(4))^2+(y(3))^2=125.
billyjean
 one year ago
what is the equation of the circle with center 4,3 passes through the point 6,2?The answer choices are A(x4)^2+(y3)^2=25,B(x4)^2+(y3)^2=125,C(x(4))^2+(y(3))^2=25 or D(x(4))^2+(y(3))^2=125.

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mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Hint: Equation of a circle with radius r and centre O(x0,y0) is \((xx0)^2+(yy0)^2=r^2\) Example: A circle with radius 6 with centre O(2,3) has the equation \((x(2))^2+(y3)^2=6^2\) or \((x+2)^2+(y3)^2=6^2\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Additional 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
 one year ago
Best ResponseYou've already chosen the best response.0the 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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