Find the equation of the circle passing through the coordinate origin and point P (1,1), and the center of the circle on the line 2x + 3Y + 1 = 0 How do you work out the equation of the vertical bisector?

If the equation of the straight line where the vertical bisector of the line segment passing through the origin (0,0) and P (1,1) is x + y = 1, then the intersection of the line and 2x + 3Y + 1 = 0 is the center of the circle. If the center of the circle is (4, - 3), then the radius of the circle is r = 5, then the equation of the circle is (x-4) &# 178; + (y + 3) &# 178; = 25
If the midpoint of O (0,0) and P (1,1) is (1 / 2,1 / 2), and the slope of the straight line OP is 1, then the vertical bisector passes through the point (1 / 2,1 / 2) and the slope is - 1, that is, y = - (x-1 / 2) + 1 / 2, and the simplification is: x + y = 1

RELATED INFORMATIONS

1. If the circle C passes through the coordinate origin and the center of the circle is on the straight line y = - 2x + 3, the equation of the circle with the minimum radius is obtained
2. It is known that a circle passes through the coordinate origin and point P (1,1), and the center of the circle is on the line 2x + 3Y + 1 = 0
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