How can the circles x^2 + y^2 = 6 and x^2 + 4y^2 = 9 have more than two points of intersection.
You need to remember the condition when two circles are coincident: the distance between the centers of the circles is zero and the radii are equal. Whee two circles are coincident then the number of solutions that check the equations of circles is infinite.
In this case, the number of solution could be four, hence there are four points of intersection.
I'll solve the system of equations to prove to you that circles intersect each other at more than two points.
You should write `x^2 ` in terms of`y^2` , hence: `x^2 = 6 - y^2`
Substituting `6 - y^2` for `x^2` in the next equation yields:
`6 - y^2 + 4y^2 = 9 =gt 6+ 3y^2 = 9`
Dividing by 3 both sides yields:
`2 +y^2 = 3 =gt y^2 = 3 - 2 =gt y^2 = 1 =gt y_(1,2) =+-1`
Substituting`y^2 = 1` for x yields:
`x^2 = 6 - 1 =gt x^2 = 5 =gt x_(1,2) = +-sqrt5`
You need to find how many possibilities of two points of intersection out of four you may get:
`C_4^2 = (4!)/(2!*2!) = 6`
Hence, there are 6 possible combinations of two points out of the four pairs of points`(1,sqrt5);(-1,sqrt5);(1,-sqrt5);(-1,-sqrt5) ` where the circles may intersect each other.
You have asked how the circles x^2 + y^2 = 6 and x^2 + 4y^2 = 9 have more than two points of intersection.
Two circles can interact with each other in 4 different ways. They do not touch each other, touch each other at a single point, intersect each other at 2 points or are coincident.
For the curves you have provided x^2 + y^2 = 6 and x^2 + 4y^2 = 9, the first curve x^2 + y^2 = 6 is a circle with center at (0, 0) and radius `sqrt 6` . x^2 + 4y^2 = 9 on the other hand is not the equation of a circle but the equation of an ellipse. In the general form of an ellipse it is represented as `(x/3)^2 + (y/(3/2))^2 = 1`
The curves x^2 + y^2 = 6 and x^2 + 4y^2 = 9 have more than 2, (exactly 4) points of intersection as x^2 + 4y^2 = 9 is the equation of an ellipse.