A circle has equation (x−9)2 + (y+2)2 = 4
A square has vertices at (1, 0), (1, 2), (−1, 2) and (−1, 0).
A straight line bisects both the area of the circle and the area of the square.
What is the x-coordinate of the point where this straight line meets the x-axis?
- A2
- B3
- C4
- D4.5
- E5
- F6
- GThe straight line is not uniquely determined by the information given, so there is more than one possible point of intersection.
- HThere is no straight line that bisects both the area of the circle and the area of the square.
Show the answer and worked solution
answer · B
- A2
- B3
- C4
- D4.5
- E5
- F6
- GThe straight line is not uniquely determined by the information given, so there is more than one possible point of intersection.
- HThere is no straight line that bisects both the area of the circle and the area of the square.
Both shapes have a centre of symmetry, and a straight line halves such a shape's area exactly when it passes through that centre. The circle's centre is (9, −2) and the square's centre is (0, 1), so the line is the one joining them — unique, since the centres are distinct. Its gradient is −2−19−0 = −13, giving y = 1 − x3, which meets the x-axis at x = 3.