rommysingh wrote:In the figure above, the circles are centered at O(0, 0) and P(10, 0). Line AB is tangent to both circles, at points A and B respectively, and intersects the x-axis at point X. What is the x-coordinate of point X ?
(1) The area of the circle centered at point P is 4 times the area of the circle centered at point O.
(2) BX is twice as long as AX.
Solutions to circle problems often require the DRAWING OF RADII.
Here, if we draw radii OA and BP, the following figure is yielded:
In the figure above:
Since vertical angles are equal, the two angles labeled "e" are equal.
Since a radius drawn to a tangent line forms a right angle, ∠OAX = ∠XBP = 90.
Since the two angles labeled "e" are equal, and ∠OAX = ∠XBP = 90, the remaining angles in the two triangles -- the two angles labeled "d" -- must also be equal.
Thus, triangle OAX and triangle XBP have the SAME COMBINATION OF ANGLES:
d-e-90.
Triangles with the same combination of angles are SIMILAR.
In similar triangles, sides opposite equal angles are in the SAME RATIO.
Thus:
BP/OA = BX/AX = PX/OX.
Statement 2:
Since BP/OA = BX/AX = PX/OX, and BX = 2(AX), we get:
PX = 2(OX).
Since OP=10, PX + OX = 10.
Substituting PX = 2(OX) into PX + OX = 10, we get;
2(OX) + OX = 10
3(OX) = 10
OX = 10/3.
Thus, the x-cooridinate of X = 10/3.
SUFFICIENT.
Statement 1:
Thus:
π(BP)² = 4 * π(OA)²
BP² = 4(OA)²
BP = 2(OA), implying that PX = 2(OX).
Statement 1 implies the same information as Statement 2.
Since Statement 2 is sufficient, Statement 1 must also be SUFFICIENT.
The correct answer is
D.
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