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In circle and line, I beleieve answer could be just 1(tangent) or 2(line). I am not able to recollect what was the method to know if line is tanget to circle or not. Anyone?
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In circle and line, I beleieve answer could be just 1(tangent) or 2(line). I am not able to recollect what was the method to know if line is tanget to circle or not. Anyone?

Actually there is a third case: when circle and line don't have any points of intersection. As for the solutions: well if it's an easy case, for example if line is \(y=2\) we can say that it's tangent to the circle right away or if line is \(y=5\) we can say right away that they don' intersect at all.

For harder cases you can use the approach used in initial post: \(y=x-2\), substitute \(x\) by \(y\) (or vise-versa) in \(x^2+y^2=4\) and then solve for \(y\) (this value will be \(y\) coordinate of the intersection point(s)). If you'll get one solution for \(y\) it would mean that line is tangent to circle (as you'll get one point (x,y)), if you'll get two solutions for \(y\) it would mean that line has two intersection points with circle (as you'll get two points (x,y)) and if you'll get no solution for \(y\) it would mean that line has no intersection point with circle.

Hope it's clear.
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I think this should be included in the circles chapter of the math book
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Hi All,

If you want to do the work, then the two equations in this question can be physically graphed, so that you can "see" the points of intersection. You don't need to graph the equations to answer this question, but you have to recognize the rarer graphing rule in this question:

The equation X^2 + Y^2 = 4 is a circle, centered around the Origin, with a radius of 2. The points (2,0), (0,2), (-2,0) and (0,-2) are all on the circumference of the circle

The equation X + Y = 2 can be rewritten as Y = -X + 2 and will be a straight line.

By definition, the line will intersect with the circle at either 0, 1 or 2 points.

Knowing the above 4 "obvious" points on the circle will allow you to quickly find 2 points on the line: (2,0) and (0,2). Then you can stop; a line can't intercept a circle at more than 2 points.

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did not understand the explanation.

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

0
1
2
3
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).

Can you please specify what part of the solution didn't you understand?

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

Answer: C (2).

how should we know what kind of shape an equation is?
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seekmba
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

did not understand the explanation.

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).

The curve x^2 + y^2 = 4 is a circle with center at (0, 0) and with a radius of 2. Since a line and a circle can intersect at 0, 1, or 2 points, we immediately eliminate D and E.

The best approach to solving this question is to draw the given circle and the line. When we do that, we will notice that both intercepts of x + y = 2 (which are (0, 2) and (2, 0)) are contained in the given circle, which means the line and the circle intersect at two points.

Though time consuming, it is also possible solve this question using algebra. Let's write y = 2 - x and substitute in the equation x^2 + y^2 = 4:

\(\Rightarrow\) x^2 + y^2 = 4

\(\Rightarrow\) x^2 + (2 - x)^2 = 4

\(\Rightarrow\) x^2 + 4 - 4x + x^2 = 4

\(\Rightarrow\) 2x^2 - 4x = 0

\(\Rightarrow\) x^2 - 2x = 0

\(\Rightarrow\) x = 2 or x = 0

Substituting x = 2 in x + y = 2 yields y = 0, so we get the point (2, 0). Similarly, substituting x = 0 in x + y = 2 yields y = 2, so we get the point (0, 2). So the line and the circle intersect in two points.

Answer: C
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Is it possible that a line can intersect at more than two points?
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