Given the equation `x^2+(y+4)^2 = 9`, find the center and radius. Then find the x and y intercepts.

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txmedteach | High School Teacher | (Level 3) Associate Educator

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To find the center of the circle, we have to recall the standard equation of a circle:

`(x-h)^2 + (y - k)^2 = r^2`

where (h,k) is the center of the circle and r is the radius.

The center is clear from the given equation: (0, -4)

To find the radius, we simply take the squre root of 9:

Radius = 3

Finding the intercepts is a bit more complicated.

Recall that at an x-intercept, the equation crosses the x-axis. In order for the graph to cross the x-axis, the y-value must be 0. Thus, we can find the x-intercepts by setting y to 0 and solving the equation for possible values of x:

`x^2+(0+4)^2 = 9`

Simplifying:

`x^2 + 16 = 9`

`x^2 = -7`

Now, we come up on a problem. In order for `x^2` to be negative, x must be an imaginary number! If the only solutions for x are imaginary, there are no real roots. This means that there are no real x-intercepts.

No, we'll do effectively the same thing to find the y-intercepts. Now, x = 0, so we'll solve the equation for y:

`0^2 + (y+4)^2 = 9`

`(y+4)^2 = 9`

Now, we can take the square root of both sides (but we'll take the positive and negative root of 9:

`y+4 = +-3`

`y = -4+-3`

`y=-1` or `y=-7`

These two vaules give us our y-intercepts: (0,-1) and (0,-7)

Now, we can test our values by graphing the equation:

Looks like we got everything right! I hope that helps!

Sources:

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