Question

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

Answer 1

(A)
`x^2+y^2-10x-12y+60=0`

Explanation:

The standard form of the equation of a circle is given as:

`(x-5)^2+(y-6)^2=1`

Simplifying the above given equation, we get

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

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

`x^2+y^2-10x-12y+61=1`

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

which is the required general form of the equation.

Hence, option A is correct.