Find the equation of the circle whose center is (1, -2) and whose graph contains the point (4, -2).
So I drew a graph, and found that the radius was 3. Then I just wrote the equation:
(x - 1)^2 + (y+2)^2 = 9 . . . is that right?
So I drew a graph, and found that the radius was 3. Then I just wrote the equation:
(x - 1)^2 + (y+2)^2 = 9 . . . is that right?
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(x - 1)^2 + (y + 2)^2 = r^2
(4 - 1)^2 + (-2 + 2)^2 = r^2
(3)^2 + 0^2 = r^2
9 = r^2
(x - 1)^2 + (y + 2)^2 = 9
You got it! Great job!
(4 - 1)^2 + (-2 + 2)^2 = r^2
(3)^2 + 0^2 = r^2
9 = r^2
(x - 1)^2 + (y + 2)^2 = 9
You got it! Great job!
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your answer is correct.
but insted of the drawing you can find the raduis by the equ of "distanse between two pts" :
{ (4-1)^2+(-2- -2)^2 }^0.5 = 3
but insted of the drawing you can find the raduis by the equ of "distanse between two pts" :
{ (4-1)^2+(-2- -2)^2 }^0.5 = 3
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Yes, but if you can't tell the radius from your graph, use the distance formula to find r.
Radius = distance from any point on the circle to the center.
Radius = distance from any point on the circle to the center.
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I'm not sure what the "right" way (the way expected by your teacher) is, but your answer is correct.