seekmba wrote:
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).