2 Points
What is the center of the circle given by the equation (x+ 5)^2 + (y-8)^2 = 1?

(- 5, 8 )

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

(x + 5)² + (y - 8)² = 1 ← is in standard form

with centre (h, k ) = (- 5, 8 )

x+y=90

x=30-y=x

step-by-step explanation:

if you understand you will get the answer

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)