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 [tex](x-x_C)^2 + (y - y_C)^2 = r^2[/tex] where [tex]C(x_C; y_C)[/tex] 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 [tex]2^2+3^2-2\cdot 2 -4\cdot 3 + 4 = 1 \neq 0[/tex], (3, 1) isn’t on it either because [tex]3^2 + 1^2 - 2\cdot 3 - 4\cdot 1 + 4 = 4 \neq 0[/tex].
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 [tex]r^2[/tex]. Thus it’s 0 if the point is on the circumference, negative if inside and positive if outside. You get [tex]3^2 + 4^2 - 6\cdot 3 -2\cdot 4 + 8 = 7[/tex], positive, so it’s outside the circle.
1a: see definition from 2a. Here, [tex]r^2 = 9[/tex].
4o: insert the y from the straight line equation (r) (which can be equivalently transformed to [tex]y = x+1[/tex]) 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. [tex]x^2+(x+1)^2+2x - 4(x+1) = x^2 + x^2 + 2x + 1 + 2x - 4x + 4 = 2x^2 + 5 = 0 \iff x^2 = 2.5 \iff x \in \{-\sqrt{2.5}, \sqrt{2.5}\}[/tex] There are two solutions, so it’s a secant.
