Find an equation of the circle and sketch it: c Radius 4, tangent to x-axis; contain (–5,8)

Answer 1

Since the circle is tangent to the x axis and the radius is 4 the center must lie on either y=4 or y=-4.

The point (-5,8) tells us the circle is above the x axis so the y portion of the coordinate =4

Finally since (-5,8) is 8 units from the x axis and the diameter is 8 (twice the radius), we know that x=-5 must be the x value of the center coordinate.

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

so

(x+5)^2+(y-4)^2= 16

I hope that helps!

Answer 2

The equation of a circle of radius r centered at (h,k) is

(x-h)²+(y-k)²=r²

So let's fill in what we know:

(x-h)²+(y-k)²=4²

Since it is tangent to the x-axis, y=0 at that point.

Also, since it has the point (-5,8) on its perimeter, the circle is above the x-axis (if it goes as high as y=8 and it's radius is only 4, then the circle must be above the x-axis). But if it is tangent to the x-axis, AND its diameter is 8, then (-5,8) MUST be at the highest point. So we know the center is at (-5,4).

So since we now have the center (h,k)=(-5,4),

(x--5)²+(y-4)²=4²

(x+5)²+(y-4)²=4²

Check: Is the equation true for x=-5 and y=8?

(-5+5)²+(8-4)²=0+4²=16 OK!

Is it true of x=-5, y=0?

(-5+5)²+(0-4)²=16 OK!

Solution: (x+5)²+(y-4)²=16