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Nikolay [14]
3 years ago
10

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

Mathematics
2 answers:
____ [38]3 years ago
8 0
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}
mash [69]3 years ago
6 0

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

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