What is the General Equation of the Circle if its center is on x+2y=3 and it is tangent to -x+y=-4 at (3,-1)?
The center of the circle is also location on a line perpendicular to -x+y=-4 that passes through (3,-1). Let's get that line:
If two lines perpendicular, then their slopes are opposite reciprocals:
-x+y=-4
y = x + 4
The slope of this line is 1. Therefore the slope of the line we are looking for is m = -1. using point slope form:
y +1 = -(x - 3)
y +1 = -x + 3
x + y = 2
The center of the circle is the intersection of x + y = 2 and x + 2y = 3. Using solution for system of equations, you can use elimination by multiplying the first equation by -1:
-x - y = -2
x + 2y = 3
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y = 1
x + 1 = 2
x = 1
Therefore, the center is located at (1,1). Next is to find the radius length which is the distance from (1,1) to (3,-1) using distance formula:
r2 = (1-3)2 + (1+1)2
r2 = (-2)2 + (2)2
r2 = 8
The equation of the circle in general form with center (1,1) and radius = √8 is:
(x-1)2 + (y-1)2 = 8
