Question 1020921: A circle whose centre lies on the line y=3x-1, touches both axes. Find the equation of the circle, which I found to be 4x^2+4y^2-4x-4y+1=0 and find the points where the circle intersects the line 2y+x=1.
Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website! .
A circle whose centre lies on the line y=3x-1, touches both axes. Find the equation of the circle,
which I found to be 4x^2+4y^2-4x-4y+1=0 and find the points where the circle intersects the line 2y+x=1.
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1. I confirm that you correctly determined the equation of the circle.
2. To find intersection points, express x = 1 - 2y from the equation of the straight line, and substitute it into the circle equation.
You will get a single quadratic equation for x.
Its solution will give you the x-coordinates of the intersection points.
When you get x-coordinates, you can restore y-coordinates using the same equation 2y+x=1 of the straight line.
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