SOLUTION: I have been struggling to figure out this math problem and I was wondering if someone could help me? Please and Thank You!! I would deeply appreciate it!! Circles Find an equat

Question 190862This question is from textbook Algebra and Trigonometry Structure and Method book 2
: I have been struggling to figure out this math problem and I was wondering if someone could help me? Please and Thank You!! I would deeply appreciate it!!
Circles
Find an equation of the circle described.
Center on line x+y=4; tangent to both coordinate axes. This question is from textbook Algebra and Trigonometry Structure and Method book 2

Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Find an equation of the circle described.
Center on line x+y=4; tangent to both coordinate axes.
.
Draw a diagram of the problem on graph paper. It'll help you see the problem.
Start by rearranging:
x+y=4
into the slope-intercept form:
y = mx + b
where
m is slope
b is y-intercept
.
x+y=4
y = -x + 4
We now KNOW that the y-intercept is at (0,4).
To find the x-intercept, set y=0 and solve for x:
y = -x + 4
0 = -x + 4
-4 = -x
4 = x
The x-intercept is at (4,0)
.
The mid-point between these two points has to be the center.
((x2+x1)/2 , (y2+y1)/2)
((4+0)/2 , (0+4)/2)
(4/2 , 4/2)
(2 , 2)
.
The point at which the center (2,2) is perpendicular to the axis is where the circle has to be tangent to... Therefore, the radius is 2.
.
Recapping:
r = 2
center at (2,2)
.
Standard circle equation is:
(x-h)^2 + (y-k)^2=r^2
where
(h,k) is the center
and
r is the radius
.
Plugging in our info:
(x-h)^2 + (y-k)^2=r^2
(x-2)^2 + (y-2)^2=2^2
(x-2)^2 + (y-2)^2=4 (this is what they're looking for)

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