brownluella
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Do the graph of (x+2)squared +(y-1)squared=4y intersect?
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Moved - Geometry question about graph and lntersection
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Deleted member 4993
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Do the graph of (x+2)squared +(y-1)squared=4y intersect?
Please share your work with us ...even if you know it is wrong
If you are stuck at the beginning tell us and we'll start with the definitions.
You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:
http://www.freemathhelp.com/forum/th...Before-Posting
Hint: You do realize that the above is an equation of a circle.
What happens to the slope/s when a graph intersects itself?
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Is there a typo or something. It appears as though you have the single equationDo the graph of (x+2)squared +(y-1)squared=4y intersect?
(x+2)2 + (y-1)2 = 4y
which is a circle.
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Deleted member 4993
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(x+2)2 +(y-1)2 = 4y → (x+2)2 +(y-3)2 = 8 → r = √8
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Deleted member 4993
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Since r^2 = 4y, then y > 0, so circle above x-axis: right?
Nope ... to the left of y-axis .....
Yes. As pointed out by Subhotosh Khan, the circle isSince r^2 = 4y, then y > 0, so circle above x-axis: right?
(x+2)2 +(y-3)2 = 8
which has a center in the second quadrant (above the x axis and to the left of the y axis), lies entirely above the x axis (only positive y allowed since \(\displaystyle 3 \gt \sqrt{8}\)) and in both the second and third quadrant (both positive and negative x is allowed since \(\displaystyle 2 \lt \sqrt{8}\)).