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1. The center of a circle is located at (5, −5) . The radius of the circle is 3.



What is the equation of the circle in general form?


​ x^2+y^2+10x−10y+47=0 ​

​ x^2+y^2+10x−10y+41=0 ​

​ x^2+y^2−10x+10y+47=0 ​

​ x^2+y^2−10x+10y+41=0 ​


2.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?


x^2+y^2−10x−12y−62=0

x^2+y^2+10x+12y+62=0

x^2+y^2−10x−12y+60=0

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

Respuesta :

caylus
Hello,

1) Answer D since (x-5)²+(y+5)²=9 ==>x²+y²-10x+10y+41=0

2) Answer C since (x-5)²+(y-6)²=1 ==>x²+y²-10x-12y+60=0

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