Asked by mally
Write down the equation of the circumscribed circle that passes through the vertices of the triangle created by line 3x − y + 6 = 0, x-axis and y-axis.
Answers
Answered by
oobleck
The line has intercepts of (-2,0) and (0,6)
The circumcenter of a right triangle lies at the midpoint of its hypotenuse. In this case, at (-1,3)
So (x+1)^2 + (y-3)^2 = 10
The circumcenter of a right triangle lies at the midpoint of its hypotenuse. In this case, at (-1,3)
So (x+1)^2 + (y-3)^2 = 10
Answered by
mathhelper
The intercepts of your slant line are (-2,0) and (0,6)
since the centre of the circle lies on the right bisectors of these it is
easy to see that the centre of the circle must be (-1,3)
distance to one end point, I picked (0,0) is √10
equation:
(x+1)^2 + (y-3)^2 = 10
since the centre of the circle lies on the right bisectors of these it is
easy to see that the centre of the circle must be (-1,3)
distance to one end point, I picked (0,0) is √10
equation:
(x+1)^2 + (y-3)^2 = 10
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