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# Does curve (x-a)^2 + (y-b)^2 = 16 intersect the Y axis? 1.

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CEO
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Does curve (x-a)^2 + (y-b)^2 = 16 intersect the Y axis? 1. [#permalink]

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28 Nov 2007, 13:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Does curve (x-a)^2 + (y-b)^2 = 16 intersect the Y axis?

1. a^2 + b^2 > 16
2. a = |b| + 5
CEO
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28 Nov 2007, 14:05
B.

(x-a)^2 + (y-b)^2 = 16 means circle in center (a,b) and radius of 4.

1. a^2 + b^2 > 16
a=0: intersect
b=0: the circle does not intersect
INSUFF

2. a = |b| + 5 ==> a>5>4 the circle does not intersect
SUFF
CEO
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28 Nov 2007, 23:37
walker wrote:
B.

(x-a)^2 + (y-b)^2 = 16 means circle in center (a,b) and radius of 4.

1. a^2 + b^2 > 16
a=0: intersect
b=0: the circle does not intersect
INSUFF

2. a = |b| + 5 ==> a>5>4 the circle does not intersect
SUFF

can you elaborate on 1? why do we set them to zero
CEO
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28 Nov 2007, 23:51
bmwhype2 wrote:
can you elaborate on 1? why do we set them to zero

a=0: intersect - Distance of circle center to Y-axis is minimum - 0
b=0: the circle does not intersect - Distance of circle center to Y-axis is maximum - b
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29 Nov 2007, 08:15
Walker can you explain how do you know its an equation for a circle? Are there particular equations to look for so we know if they are triangles, circles, rectangles??
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29 Nov 2007, 15:09
shubhampandey wrote:
Walker can you explain how do you know its an equation for a circle? Are there particular equations to look for so we know if they are triangles, circles, rectangles??

http://www.analyzemath.com/CircleEq/Tutorials.html
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29 Nov 2007, 15:18

we have a center of a circle - (a,b)
for any point (x,y) distance between (x,y) and the center must be constant (r -radius).

So, r^2=(x-a)^2+(y-a)^2 for any point on the circle. It is simply Pythagorean theorem.
29 Nov 2007, 15:18
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