How do you write the equation of the circle whose centre is at (-5, 3) and which passes through the point (-4, -5)?

1 Answer
Nov 3, 2017

Answer:

#(x+5)^2+(y-3)^2=65#

Explanation:

The general equation for a circle with center #(a,b)# and radius #r# is
#color(white)("XXX")(x-a)^2+(y-b)^2=r^2#

We are given
#color(white)("XXX")a=-5# and
#color(white)("XXX")b=+3#

so we only need to find the radius.
The radius is the distance from the center to any point on the circumference.
Given the center, #(-5,3)#, and a point on the circumference, #(-4,-5)# we can evaluate the radius using the Pythagorean Theorem
#color(white)("XXX")r^2=(-5-(-4))^2+(3-(-5))^2#

#color(white)("XXX=")=(-1)^2+8^2#

#color(white)("XXX=")=65#

Therefore the equation of the circle is
#color(white)("XXX")(x-(-5))^2+(y-3)^2=65#
or, simplifying the first term
#color(white)("XXX")(x+5)^2+(y-3)^2=65#