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Does the point (3, -1) lie on the circle (x+1)^2+(y+1)^2=16?A)no; when you plug in the point for x and y in the equation, you do not get a true statementB)no; the point is not represented by (h, k) in the equationC)yes; the point is represented by (h, k) in the equationD)yes; when you plug the point in for x and y you get a true statement

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SOLUTION

The given equation is:

[tex]\left(x+1\right)^2+\left(y+1\right)^2=16[/tex]

The graph of the circle is shown:

Notice that the point (3,-1) lies on the circumference of the circle hence the point is not the center of the circle

Therefore the required answer is:

Yes; when you plug the point in for x and y you get a true statement

Ver imagen NethraP293498

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