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Quick question, since I haven't been able to find the answer throughout the forum. When question says that circle is Xssqrd+Ysqrd=1, does that mean that radius of a circle on an XY plane is 1?
Appreciate the help guys!
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Re: Cricle in a plane
[#permalink]
17 Dec 2010, 09:54
1
Kudos
Expert Reply
MisterEko wrote:
Quick question, since I haven't been able to find the answer throughout the forum. When question says that circle is Xssqrd+Ysqrd=1, does that mean that radius of a circle on an XY plane is 1?
Appreciate the help guys!
THEORY: In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((x-a)^2+(y-b)^2=r^2\)
This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.
If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\)
So, if you have x^2+y^2=1 then you know that this circle is centered at the origin and has the radius equal to \(\sqrt{1}=1\)
Re: Cricle in a plane
[#permalink]
17 Dec 2010, 13:30
Bunuel wrote:
MisterEko wrote:
Quick question, since I haven't been able to find the answer throughout the forum. When question says that circle is Xssqrd+Ysqrd=1, does that mean that radius of a circle on an XY plane is 1?
Appreciate the help guys!
THEORY: In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((x-a)^2+(y-b)^2=r^2\)
This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.
If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\)
So, if you have x^2+y^2=1 then you know that this circle is centered at the origin and has the radius equal to \(\sqrt{1}=1\)
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.