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Question 1088896: Given circle x^2-2x+y^2-16y+39=0 with center A and y-intercepts B(0;p) and C(0;q) where p <
q
1 determine the values of p and q
2 show that point D(2;13) lies on circumference of the circle
3 prove that point B,A and D are collinear
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x^2-2x+y^2-16y+39=0
To get the equation of a circle, one needs to write this as (x-a)^2+(y-b)^2=r^2
The first thing is to move the constant to the right side.
Then rewrite as x^2-2x+y^2-16y=-39
One needs to complete the square of x^2=2x. That is half the -2, squared, which is +1; x^2-2x+1. One has to add 1 to the other side to balance the equation.
Do the same with the y, by taking half of -16, or -8, and squaring it to get 64.
complete the square of x and y to get x^2-2x+1+y^2-16x+64=-39+1+64
This can be written as the sum of two squares equalling a constant, which is the radius squared.
(x-1)^2+(y-8)^2=26
center is at (1, 8) and radius is sqrt(26). One takes the opposite sign of the x-1, and the center has x coordinate 1. The y-coordinate of the center is 8
y-intercept is where x=0, so one puts a 0 into any place there is an x. One could do it at the beginning as well and get y^2-16y+39=0; that is y^2-16y=-39, and completing the square y^2-16y+64=25. That is (y-8)^2=5^2
1+(y-8)^2=26
(y-8)^2=25
y=8+/- 5 or 3 and 13. p is 3 and q is 13.
center is (1, 8) and B is (0, 3) and D is (2, 13). If these points are collinear, the slope between any two of them must be the same. One does the slope formula for them.
slope between center and B is -5/-1 or 5
slope between B and D is 10/2=5
check to see if (2, 13) is a solution of the circle equation by substituting (2, 13) into the equation for x and y respectively: 1^2+5^2=26, and that is true.
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