Monte-Carlo estimating circle

Question

suppose that the equation of a circle is

(x - 3)^2 + (y + 2)^2 = 16

A) define the corresponding distribution f(x) and f(y) and show how a sample point (x,y) is determined using the (0,1) random pair R1 , R2

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Answer 1

it's circle with radius = sqr16 = 4

and center (3,-2)

general formula for a circle is

(x-h)^2 + (y-k)= r^2 where (h,k) is the center and r= the radius

(x-3)^2 + (y+2)^2= 16

(y+2)^2 =16-(x-3)^2

y+2 = sqr(16-(x-3)^2)

y = f(x) = -2 +/- sqr[16-x^2 -6x +9]

x= f(y) = 3 +/-sqr[16-y^2 +4y +4]

(3, 2) is on the circle, the top of the circle

(3,-4) is the bottom of the circle

each directly above and below the center

the range is y= 2 to -4

left most point of the circle is (-1,2) and

the right most point is (7,2) directly left and right of the center

the domain is x= -1 to 7

pick any x value in that domain to find 2 y values

pick any y value in that range to find 2 x values

pick any x or y that's an endpoint of the doman or range and there is just one corresponding y or x coordinate