Asked by mumba
the equation of the circle is x^2+y^2+2x-8y-1=0.
Find the coordinates of the point where the circle crosses the x-axis
Find the coordinates of the point where the circle crosses the x-axis
Answers
Answered by
drwls
y will equal 0 where the circle crosses the x axis.
x^2 +2x -1 = 0
x = (1/2)[-2 +/-sqrt8]
= -1 +/-sqrt2
= -2.414 and 0.414
x^2 +2x -1 = 0
x = (1/2)[-2 +/-sqrt8]
= -1 +/-sqrt2
= -2.414 and 0.414