Answer:
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 <em>r</em> is the radius. (a) and (d) can’t be it because they contain non-zero factors on <em>xy</em>. (b) isn’t an equation.
5o: just put the given (<em>x</em>, <em>y</em>) 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 (<em>x</em>, <em>y</em>) from the given point <em>M</em>. 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.