How do you write the equation for a circle with center at (2,-5) and passing through (3,4)?

1 Answer
Jun 10, 2016

Answer:

#(x-2)^2+(y+5)^2=21#

Explanation:

The general equation of a circle is:

#color(blue)(|bar(ul(color(white)(a/a)(x-h)^2+(y-k)^2=r^2color(white)(a/a)|)))#

where:
#x=#x-coordinate
#h=#x-coordinate of centre
#y=#y-coordinate
#k=#y-coordinate of centre
#r=#radius

Start by plugging #(h,k) = (2,-5)# into the equation.

#(x-(2))^2+(y-(-5))^2=r^2#

Replace #(x,y)# with #(3,4)#.

#(3-2)^2+(4-(-5))^2=r^2#

Simplify.

#(1)^2+(9)^2=r^2#

#1+81=r^2#

#21=r^2#

Rewrite the equation by including the centre and the radius.

#(x-2)^2+(y-(-5))^2=21#

#color(green)(|bar(ul(color(white)(a/a)color(black)((x-2)^2+(y+5)^2=21)color(white)(a/a)|)))#