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Sagot :
REMEMBER that if the points lie on the circle, it would have the same distance from the center then we can say that if points (-2,5), (-2,-1) and (4,-1) are having the same distance from the center located at (1,2) then the points are part of the circle.
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Distance Formula:
[tex]d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]
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Distance from point (-2,5) to the center (1,2)
[tex]d = \sqrt{(1-(-2))^2+(2-5)^2}[/tex]
[tex]d = \sqrt{3^2+(-3)^2}[/tex]
[tex]d = 3\sqrt2[/tex]
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Distance from the point (-2,-1) to the center (1,2)
[tex]d = \sqrt{(1-(-2))^2+(2-(-1))^2}[/tex]
[tex]d = \sqrt{3^2 + 3^2}[/tex]
[tex]d = 3\sqrt2[/tex]
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Distance from the point (4,-1) to the center (1,2)
[tex]d = \sqrt{(1-4)^2+(2-(-1))^2}[/tex]
[tex]d = \sqrt{(-3)^2+3^2}[/tex]
[tex]d = 3\sqrt2[/tex]
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Since the distances of the points to the center is equal then we can say that the points lie on the circle with the radius of 3√2.
---------------------------------
Distance Formula:
[tex]d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]
--------------------------------
Distance from point (-2,5) to the center (1,2)
[tex]d = \sqrt{(1-(-2))^2+(2-5)^2}[/tex]
[tex]d = \sqrt{3^2+(-3)^2}[/tex]
[tex]d = 3\sqrt2[/tex]
--------------------------------
Distance from the point (-2,-1) to the center (1,2)
[tex]d = \sqrt{(1-(-2))^2+(2-(-1))^2}[/tex]
[tex]d = \sqrt{3^2 + 3^2}[/tex]
[tex]d = 3\sqrt2[/tex]
-------------------------------
Distance from the point (4,-1) to the center (1,2)
[tex]d = \sqrt{(1-4)^2+(2-(-1))^2}[/tex]
[tex]d = \sqrt{(-3)^2+3^2}[/tex]
[tex]d = 3\sqrt2[/tex]
------------------------------
Since the distances of the points to the center is equal then we can say that the points lie on the circle with the radius of 3√2.
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