PDA

View Full Version : Find the equation of the circle

Sara33
06-01-2011, 03:16 AM
A circle with its centre on the y-axis intersects the graph of y x = at the origin, O, and exactly two other distinct points, A and B. Prove that the ratio
of the area of triangle ABO to the area of the circle is always 1 : ?.

Since the circle has centre on the y-axis (say, has coordinates (0,b)), then its radius is equal to b (and b must be positive for there to be three points of intersection). So the circle has equation x^2+(y-b)^2=b^2 I do not understand the sentence in bold. Can you please explain?

tkhunny
06-01-2011, 07:34 AM
1) It is worded badly. The relation x*y = ?? isn't complete and it doesn't go through the origin.
2) Fix the relation, x*y = ?? We've a value missing.
3) Fix the question. Passes through the origin and it intersects x*y = ?? in two places.

Sara33
06-02-2011, 04:51 AM
A circle with its centre on the y-axis intersects the graph of y=[x] (these square brackets are suppose to be straight lines) at the origin, O, and exactly two other distinct points, A and B. Prove that the ratio
of the area of triangle ABO to the area of the circle is always 1 : ?.

Since the circle has centre on the y-axis (say, has coordinates (0,b)), then its radius is equal to b (and b must be positive for there to be three points of intersection). So the circle has equation x^2+(y-b)^2=b^2. I do not understand the sentence in bold. Can you please explain?

tkhunny
06-02-2011, 05:07 AM
Circle of radius 'r' centered at the Origin: x^2 + y^2 = r^2
Circle of radius 'r' centered somewhere on the positive y-axis, (0,b): x^2 + (y-b)^2 = r^2
Circle centered somewhere on the positive y-axis, (0,b), and passing through the Origin: x^2 + (y-b)^2 = b^2

If the center is (0,b) and it passes through the Origin, the radius must be 'b'.