Answer with Step-by-step explanation:
We are given that
[tex]x^2+y^2=(\sqrt 2)^2[/tex]
[tex](x-3)^2+(y-3)^2=32=(4\sqrt 2)^2[/tex]
Compare with the equation of circle
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where center of circle=(h,k)
r=Radius of circle
a.Center of circle=(0,0)
Radius=[tex]\sqrt 2[/tex] units
Center of second circle=(3,3)
Radius of second circle=[tex]4\sqrt 2[/tex] units
b.Distance formula:[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2[/tex]
Using the formula
The distance between the centers of two circle
=[tex]\sqrt{(3-0)^2+(3-0)^2}=3\sqrt 2[/tex]
Hence, the distance between the centers of two circle =[tex]3\sqrt 2[/tex] units.
c.
Substitute x=-1 and -1
[tex]1+1=2=[/tex]
[tex](1-3)^2+(1-3)^2=32[/tex]
The circle must be tangent because there is just one point (-1,-1) is common in both circles and satisfied the equations of circle.
