Will award brainliest!

write the standard form equation for the circle whose center is at (0, 0) and that is tangent to the line x + y = 8.

a). x^2 + y^2 - 8 = 0
b). x^2 + y^2 - 32 = 0
c). x^2 + y^2 - 64 = 0

B). x^2 + y^2 - 32 = 0

Step-by-step explanation:

If it is tangent to x+y = 8, then the distance from (h,k) to a point on that line  can be found by using

    d = |A*h+B*k+C| / sqrt(A^2+B^2)

Where Ax + By +c = 0 is the equation of the line and (h,k) is the center of the circle

1x+1y -8 =0

so A =1 B = 1 C = -8

h,k = 0,0

   d = |1*0+1*0+-8| / sqrt(1^2+1^2)

   d = |-8| / sqrt(2)

   d = 8 /sqrt(2)

Multiplying by the sqrt(2)/ sqrt(2) =

    d = 8 /sqrt(2) * sqrt(2)/ sqrt(2)

    d = 8 sqrt(2)/2

   d  =4 sqrt(2)

This is the radius

The standard equation for a circle is in vertex form

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

(x-0)^2 + (y-k) ^2 = (4 sqrt(2))^2

x^2 + y^2 = 16*2

x^2 + y^2 = 32

Rewriting in standard form

x^2 + y^2 -32 = 32-32

x^2 +y^2 -32 =0

han when lo ahe whn hew anh ? ?

explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.