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 = 0B). x^2 + y^2 - 32 = 0C). x^2 + y^2 - 64 = 0

Answer:B). x^2 + y^2 - 32 = 0Step-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 circle1x+1y -8 =0so A =1 B = 1 C = -8h,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 radiusThe 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))^2x^2 + y^2 = 16*2x^2 + y^2 = 32Rewriting in standard formx^2 + y^2 -32 = 32-32x^2 +y^2 -32 =0