Question

How do you write the equation for a circle with center (-1, -3) and passing through (-2, 0)?

Answer 1

`(x+1)^2+(y+3)^2=10`

Explanation:

Recall that the general formula for a circle is:

`color(blue)(|bar(ul(color(white)(a/a)(x-a)^2+(y-b)^2=r^2color(white)(a/a)|)))`

where
`x=`x-coordinate
`y=`y-coordinate
`a=`x-coordinate of circle's centre
`b=`y-coordinate of circle's centre
`r=`radius

Using the formula, substitute the circle's centre, `(-1,-3)`.

`(x+1)^2+(y+3)^2=r^2`

Plug in the point, `(-2,0)`.

`(-2+1)^2+(0+3)^2=r^2`

Solve for `r`.

`r=sqrt((-2+1)^2+(0+3)^2)`

`r=sqrt(1+9)`

`r=sqrt(10)`

Rewrite the equation.

`(x+1)^2+(y+3)^2=(sqrt(10))^2`

`color(green)(|bar(ul(color(white)(a/a)color(black)((x+1)^2+(y+3)^2=10)color(white)(a/a)|)))`