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

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CEO
Joined: 21 Jan 2007
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Does curve (x-a)^2 + (y-b)^2 = 16 intersect the Y axis? 1. [#permalink]  28 Nov 2007, 12:48
Does curve (x-a)^2 + (y-b)^2 = 16 intersect the Y axis?

1. a^2 + b^2 > 16
2. a = |b| + 5
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2143 [0], given: 359

Expert's post
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
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 349 [0], given: 4

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
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2143 [0], given: 359

Expert's post
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
Intern
Joined: 25 Nov 2007
Posts: 38
Followers: 0

Kudos [?]: 8 [0], given: 0

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??
CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 349 [0], given: 4

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
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2143 [0], given: 359

Expert's post

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.
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