URGENTE, PERGUNTAS DE MARCAR:

2a: (c)

5o: (1, 3) and (1,1)

3a: (b)

1a: (d)

4o: (b)

Step-by-step explanation:

2a: the equation of a circle circumference needs to be transformable to the form where is the center and r is the radius. (a) and (d) can’t be it because they contain non-zero factors on xy. (b) isn’t an equation.

5o: just put the given (x, y) into the equations and see if it holds. (2, 3) isn’t on the circumference of (1) because , (3, 1) isn’t on it either because .

3a: calculate the value of the left-hand side term of the equation using (x, y) from the given point M. That’s the difference of square distance to the center to the square radius . Thus it’s 0 if the point is on the circumference, negative if inside and positive if outside. You get , positive, so it’s outside the circle.

1a: see definition from 2a. Here, .

4o: insert the y from the straight line equation (r) (which can be equivalently transformed to ) into the circumference equation. If it yields no solution, that’s outside, it there’s exactly one solution, that’s a tangent and if there are two solutions, it’s a secant. There are two solutions, so it’s a secant.

i would say true? if the unit was informational and it "refreshed your understanding", then true.