Answer: (x+2)²+(y+1)²=1
Step-by-step explanation:
Circle equation
[tex]\boxed {(x-x_o)^2+(y-y^0)^2=R^2}[/tex]
O(-2,-2) A(-3,-1)
Hence,
[tex](x-(-2))^2+(y-(-1))^2=R^2\\(x+2)^2+(y+1)^2=R^2\\[/tex]
Line length formula
[tex]\boxed {L^2=(x_2-x_1)^2+((y_2-y_1)^2}[/tex]
x₁=-2 x₂=-3 y₁=-1 y₂=-1
AO²=(-3-(-2))²+(-1-(-1))²
AO²=(-3+2)²+(-1+1)²
AO²=(-1)²+0²
AO²=1+0
AO²=1
AO=R
Hence,
R²=1²
R²=1
