What is the center of a circle whose equation is x2 + y2 – 12x - 2y + 12 = 0?
(-12, -2)
O (-6, -1)
(6, 1)
O (12, 2)

Solution:

Given equation of circle is:

Move the constant to other side

Group the equation

Add 36 and 1 from both sides

Which is simplified as:

--------- eqn 1

The equation of circle is given as:

Where, r is radius and (h, k) is center

Compare eqn 1 with above general equation

(h, k) = (6, 1)

Thus the center of a circle is (6, 1)

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)

step-by-step explanation:

for the free