Question 1006639: what is the equation in general form of the circle passing through the intersection of 2x-3y+6=0 and x+3y-6=0 with center at (3,-1)? show solution.
Answer by Cromlix(4381)   (Show Source):
You can put this solution on YOUR website! Hi there,
First solve the simultaneous equations:
2x - 3y + 6 = 0
x + 3y -6 = 0
2x - 3y = -6
x + 3y = 6
Add
3x = 0
x = 0
Substitute x = 0 into:
x + 3y = 6
0 + 3y = 6
3y = 6
y = 2.
Intersection = (0,2)
Next, distance formula from (0,2)
to (3,-1)
Formula √ (x2 - x1)^2 + (y2 - y1)^2
√(3 - 0)^2 + (-1 - 2)^2
√(3)^2 + (-3)^2
√9 + 9
√18
This is the radius.
Circle Formula:
(x - a)^2 + (y - b)^2 = r^2
(x - 3)^2 + (y - (-1))^2 = (√18)^2
(x - 3)^2 + (y + 1)^2 = 18
Hope this helps :-)
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