Question

Find the equation for the circle with center (-2,-1) and passing through (-3,-1).

Answer 1

Answer: (x+2)²+(y+1)²=1

Explanation:

Circle equation

`\boxed {(x-x_o)^2+(y-y^0)^2=R^2}`

O(-2,-2) A(-3,-1)

Hence,

`(x-(-2))^2+(y-(-1))^2=R^2\\(x+2)^2+(y+1)^2=R^2\\`

Line length formula

`\boxed {L^2=(x_2-x_1)^2+((y_2-y_1)^2}`

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