The proof that point (1,root 3) lies on the circle that is centered at the origin and contains the point (0,2) is found in the table below. What is the justification for the 5th statement

[tex]The \ \mathbf{distance} \ d \ between \ the \ \mathbf{points} \ (x_{1},y_{1}) \ and \ (x_{2},y_{2}) \ in \ the \ plane \ is:\\ \\ d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]

As in the statement:

The distance form [tex](0,0)[/tex] to [tex](0,2)[/tex] is [tex]\sqrt{(0-0)^2+(2-0)^2}=\sqrt{2^2}=2[/tex] whose justification is the Distance Formula.

Here in this statement the justification is also the Distance Fromula, so we take the distance from [tex](0,0) \ to \ (1,\sqrt{2})[/tex] whose result is also 2 and is the radius of the circle.

raphealnwobi raphealnwobi · Answer 2 · 2022-03-28T11:51:25+02:00

The justification for the 5th statement is the definition of radius.

The standard equation of a circle

The standard equation of a circle is given by:

(x - h)² + (y - k)² = r²

Where (h, k) is the circle center and r is the radius.

The circle is centered at origin, hence h = 0 , k = 0.

The radius is the distance from (0, 0) to (0 , 2), hence:

[tex]Radius = \sqrt{(2-0)^2+(0-0)^2}=2[/tex]

Since the point (1, √3) lies on the circle, hence the distance from (0, 0) to (1, √3) is the radius:

[tex]Radius = \sqrt{(1-0)^2+(\sqrt{3} -0)^2}=2[/tex]

The justification for the 5th statement is the definition of radius.

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