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Bunuel
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Is there a quick way to recognize the equation of a circle?

Thanks

Yes.

In an xy 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: \(x^2+y^2=r^2\)

For more check: math-coordinate-geometry-87652.html
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Very helpful, thanks Bunuel!
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Bunuel
Official Solution:

How many points of intersection does the curve \(x^2 + y^2 = 4\) have with line \(x + y = 2\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


Curve \(x^2 + y^2 = 4\) is a circle with radius 2 and the center at the origin. A line cannot have more than 2 points of intersection with a circle, so we can eliminate choices D and E. To answer the question, we can either draw a sketch or solve the system of equations. The second approach gives \((2 - y)^2 + y^2 = 4\) or \(2y^2 - 4y + 4 = 4\) or \(y^2 - 2y = 0\) from where \(y = 0\) and \(y = 2\). Thus, the line and the circle intersect at points \((2, 0)\) and \((0, 2)\).

Alternative Explanation

Look at the diagram below:



\(x^2 + y^2 = 4\) is the equation of a circle centered at the origin and with the radius of 2. Now, \(y=2-x\) is the equation of a line with \(x\)-intercept \((2, 0)\) and \(y\)-intercept \((0,2)\). Notice that these points are also the intercepts of given circle with \(X\) and \(Y\) axis hence at these points the line and the circle intersect each other.


Answer: C
Bunuel I did not understand this part "\((2 - y)^2 + y^2 = 4\) or \(2y^2 - 4y + 4 = 4\) or \(y^2 - 2y = 0\) from where \(y = 0\) and \(y = 2\). Thus, the line and the circle intersect at points \((2, 0)\) and \((0, 2)\)." like where did this come from "[m](2 - y)^2 + y^2 = 4" ?
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Bunuel
Official Solution:

How many points of intersection does the curve \(x^2 + y^2 = 4\) have with line \(x + y = 2\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


Curve \(x^2 + y^2 = 4\) is a circle with radius 2 and the center at the origin. A line cannot have more than 2 points of intersection with a circle, so we can eliminate choices D and E. To answer the question, we can either draw a sketch or solve the system of equations. The second approach gives \((2 - y)^2 + y^2 = 4\) or \(2y^2 - 4y + 4 = 4\) or \(y^2 - 2y = 0\) from where \(y = 0\) and \(y = 2\). Thus, the line and the circle intersect at points \((2, 0)\) and \((0, 2)\).

Alternative Explanation

Look at the diagram below:



\(x^2 + y^2 = 4\) is the equation of a circle centered at the origin and with the radius of 2. Now, \(y=2-x\) is the equation of a line with \(x\)-intercept \((2, 0)\) and \(y\)-intercept \((0,2)\). Notice that these points are also the intercepts of given circle with \(X\) and \(Y\) axis hence at these points the line and the circle intersect each other.


Answer: C
Bunuel I did not understand this part "\((2 - y)^2 + y^2 = 4\) or \(2y^2 - 4y + 4 = 4\) or \(y^2 - 2y = 0\) from where \(y = 0\) and \(y = 2\). Thus, the line and the circle intersect at points \((2, 0)\) and \((0, 2)\)." like where did this come from "(2 - y)^2 + y^2 = 4" ?

First of all, check this post: https://gmatclub.com/forum/m16-184073.html#p1437884

We are given that \(x^2 + y^2 = 4\) have with line \(x + y = 2\). From \(x + y = 2\) we have \(x = 2 - y\). Substitute \(x = y - 2\) into \(x^2 + y^2 = 4\) to get \((2 - y)^2 + y^2 = 4\).
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