how do I find the intercept for x^2+(y-4)^2=0
4 answers
typo for x^2+(y-4)^2=16
That makes more sense. A circle of zero radius is hard to see.
circle center at x = 0 , y = 4
circle radius = sqrt 16 = 4
so it just grazes the x axis at x = 0
it crosses the y axis at x =0 and , y = 0 and y = 8
circle center at x = 0 , y = 4
circle radius = sqrt 16 = 4
so it just grazes the x axis at x = 0
it crosses the y axis at x =0 and , y = 0 and y = 8
how does y=8 do you know?
when x=0, (y-4)^2=16
So, y-4 = ±4
y = 4±4
y=0,8
So, y-4 = ±4
y = 4±4
y=0,8