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Given the two circles x²+y²−2x+2y+1=0,x²+y²−6x+2y+6=0, we can say that:

(a) they are tangent (one point of intersection)
(b) they intersect at four distinct points
(c) they do not intersect and the second is internal to the first
(d) they intersect at two distinct points
(e) they do not intersect and the first is internal to the second The correct answer is: they intersect at two distinct points

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Final answer:

The two circles intersect at two distinct points.

Step-by-step explanation:

The two circles x²+y²−2x+2y+1=0 and x²+y²−6x+2y+6=0 intersect at two distinct points.

To find the points of intersection, we can solve the system of equations formed by the two circle equations.

By subtracting the second equation from the first equation, we get -4x + 4x = -5, which simplifies to 0 = -5. This indicates that there are no real solutions and therefore the circles do not intersect at four distinct points or share any common points. Thus, the correct answer is that the circles intersect at two distinct points.

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