The standard form of the equation of a circle is (x−5)2+(y−6)2=1. what is the general form of the equation?
x2+y2+10x+12y+60=0
x2+y2−10x−12y−62=0
x2+y2+10x+12y+62=0
x2+y2−10x−12y+60=0

x2+y2−10x−12y+60=0

Step-by-step explanation:

The standard form of the equation of a circle is usually given in the form;

The center of the circle is (a, b) while r denotes the radius.

The general form of the equation of a circle is obtained by expanding the brackets, collecting like terms and simplifying.

The standard form in this case has been given as;

(x−5)^2+(y−6)^2=1

We simply open the brackets and simplify to obtain the general form;

(x−5)^2 = (x−5) (x−5)

            = x^2 -5x -5x +25

            = x^2 -10x +25

(y−6)^2 = (y-6) (y-6)

            = y^2 -6y - 6y +36

            = y^2 -12y +36

(x−5)^2+(y−6)^2 = x^2 -10x +25 + y^2 -12y +36 = 1

x^2 -10x +25 + y^2 -12y +36 = 1

x^2 + y^2 -10x -12y +25 + 36 = 1

x^2 + y^2 -10x -12y + 60 = 0

answer:

24pi or letter c