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

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

Answer 2

Answer:

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

Option 1 is correct.

Step-by-step 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: $x^2+y^2-6x-2y-6=0$