6
advanced-math

Solve the system: y=x2y = x^2 and x2+(y−2)2=4x^2 + (y - 2)^2 = 4.

A

(0,0),(±3,3)(0, 0), (\pm \sqrt{3}, 3)

B

(0,0)(0, 0) only

C

(±2,4)(\pm 2, 4)

D

No solution

Correct Answer: A

Choice A is the correct answer. Substitute x2=yx^2 = y into the circle equation.

  1. y+(y−2)2=4y + (y - 2)^2 = 4.
  2. y+y2−4y+4=4y + y^2 - 4y + 4 = 4.
  3. y2−3y=0⇒y(y−3)=0y^2 - 3y = 0 \Rightarrow y(y - 3) = 0.
  4. y=0y = 0 or y=3y = 3.
  5. If y=0,x2=0⇒x=0y=0, x^2=0 \Rightarrow x=0. Point (0,0)(0,0).
  6. If y=3,x2=3⇒x=±3y=3, x^2=3 \Rightarrow x=\pm \sqrt{3}. Points (3,3),(−3,3)(\sqrt{3}, 3), (-\sqrt{3}, 3).

Choice B is incorrect. Choice C is incorrect. Choice D is incorrect.

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