Answer:
[tex]$ x^2 + y^2 - 3x -y +6 = 0 $[/tex]
Step-by-step explanation:
The equation of a circle with center [tex]$ (h,k) $[/tex] and radius [tex]$ r$[/tex] is given by
[tex]$ (x - h)^2 + (y - k)^2 = r^2 $[/tex]
Here the center is: [tex]$ (3,1) $[/tex] and radius is [tex]$ 2 $[/tex].
Therefore, we have:
[tex]$ (x - 3)^2 + (y - 1)^2 = 2^2 $[/tex]
[tex]$ \implies x^2 -3x +9 +y^2 -y +1 = 4 $[/tex]
[tex]$ \implies x^2 + y^2 -3x -y +6 = 0 $[/tex]
Answer:
Step-by-step explanation:
an equation of the circle Center at the w(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : r = 2 and a =3 b =1
so : (x-3)² +(y-1)² = 2²
