Question

Find an equation of the circle that has center -(-5,6) and passes through (-1,1).

Answer 1

`(x+5)^2+(y-6)^2=41`

Explanation:

Hi there!

Equation of a circle:
`(x-h)^2+(y-k)^2=r^2` where the circle is centered at
`(h,k)` and r is the radius

1) Plug in the center (-5,6)

`(x-(-5))^2+(y-6)^2=r^2\\(x+5)^2+(y-6)^2=r^2`

2) Plug in the given point (-1,1) and solve for r²

`(-1+5)^2+(1-6)^2=r^2\\(4)^2+(-5)^2=r^2\\16+25=r^2\\41=r^2`

Therefore, r² is 41. Plug this back into
`(x+5)^2+(y-6)^2=r^2`:

`(x+5)^2+(y-6)^2=41`

I hope this helps!