Question

The standard form of the equation of a circle is (x−3)2+(y−1)2=16.

What is the general form of the equation?

Answer 1

General Form:
`x^2+y^2-6x-2y-6=0`

Option 1 is correct.

Explanation:

The standard form of the equation of a circle is (x−3)²+(y−1)²=16

The general form of circle: x²+y²+2gx+2fy+c=0

Formula:

`(a-b)^2=a^2-2ab+b^2`

`(x-3)^2+(y-1)^2=16`

`x^2-6x+9+y^2-2y+1=16`

`x^2+y^2-6x-2y-6=0`

General Form:
`x^2+y^2-6x-2y-6=0`

Standard form:
`(x-3)^2+(y-1)^2=16`

Hence, The General Form:
`x^2+y^2-6x-2y-6=0`

Answer 2

`Use:(a-b)^2=a^2-2ab+b^2\\\\(x-3)^2+(y-1)^2=16\\\\x^2-2\cdot x\cdot3+3^2+y^2-2\cdot y\cdot1+1^2=16\\\\x^2-6x+9+y^2-2y+1=16\\\\x^2+y^2-6x-2y+10=16\ \ \ |-16\\\\\boxed{x^2+y^2-6x-2y-6=0}`