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

1 Answer
Apr 30, 2016

Answer:

#(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)|)))#