## dquinao 2 years ago which is not a point on the circle x^2+(y-5)^2 =25? a.(0,5) b. (5,5) c. (5,0) d.(3,9)

1. ByteMe

plug in those points and see which one holds true

2. whpalmer4

\(x^2+y^2 = r^2\) is the equation of a circle with radius \(r\) centered at the origin. In this case, \(r^2=25\) so the radius of the circle is \(5\) because \(5^2 = 25\). When we subtract or add something to \(x\) or \( y\) before squaring it, it has the effect of shifting the circle by the amount added or subtracted along the axis corresponding to the variable where we added or subtracted. Subtracting 5 from \(y\) has the effect of shifting the circle up by 5.

3. frederickinrio

resposta = d.(3,9) and so replace the x and y

4. dquinao

5. whpalmer4

Come on, this is easy if you sketch the circle. |dw:1361068806890:dw| Now shift everything up by 5 (add 5 to the y value). Which point doesn't belong?

6. ByteMe

yes... (5, 0) is NOT in the circle: \(\large x^2+(y-5)^2 =25 \) you are correct.

7. whpalmer4

|dw:1361068902180:dw|

8. dquinao

@whpalmer4 is the answer is d?

9. whpalmer4

No, the answer is that (5,0) is a point not on the circle, as you can see from the diagram.

10. whpalmer4

(3,9) you would have to check by substituting it in the equation, but given that the format of the question tells you that only one of the points isn't on the circle, it's clear that (5,0) is the only one.

11. whpalmer4

In general, it's useful to understand how adding and subtracting inside the squared terms moves the figure around. Much easier to recognize the effect than to figure it out by laboriously plotting values! :-)

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