how do I find the intercept for x^2+(y-4)^2=0
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
how does y=8 do you know?
when x=0, (y-4)^2=16
So, y-4 = ±4
y = 4±4
y=0,8
To find the intercepts of the equation x^2 + (y - 4)^2 = 0, you first need to rewrite the equation in a more familiar form. This equation represents a circle centered at the point (0, 4) with a radius of 0.
Since the equation has a radius of 0, it means the circle is essentially a single point, which is the center of the circle: (0, 4). Therefore, this circle doesn't have any intercepts with the x or y-axis.
In general, if you have an equation in the form of (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius, you can find the intercepts by solving for x and y when either one takes on a value of 0. However, in this particular equation, since the radius is 0, there are no intercepts.