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NEED HELP QUICK! 30 POINTS!!!

Find an equation of the circle and sketch it: k Center on line y=8–x, tangent to both axes

User RazerM
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1 Answer

6 votes

Answer:


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

Explanation:

Let A(a,b) be the center of the circle. This point lies on the liny y = 8 - x.

So,


b=8-a

Let points B and C be tangent points, so


B(a,0)\\ \\C(0,b)=(0,8-a)

Find the radii AB and AC:


AB=√((a-a)^2+(8-a-0)^2)=|8-a|\\ \\AC=√((a-0)^2+(8-a-(8-a))^2)=|a|

All circle's radii are the same, so


|8-a|=|a|\\ \\8-a=a\ \text{or}\ 8-a=-a

Solve each equation:


\2a=8\\ \\a=4

or


8=0

has no solutions.

Thus, the center of the circle is at point (4,4) and the radius is r = |a| = 4.

Therefore, the equation of the circle is


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

NEED HELP QUICK! 30 POINTS!!! Find an equation of the circle and sketch it: k Center-example-1
User Vonaka
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