The point (-4,-3) lies on a circle whose center is at (0,6). How do you find an equation of this circle?

2 Answers
Feb 15, 2016

Answer:

#x^2+(y-6)^2=109#

Explanation:

If the circle has a center at #(0,6)# and #(-4,-3)# is a point on its circumference,
then it has a radius of:
#color(white)("XXX")r=sqrt((0-(-3))^2+(6-(-4))^2)=sqrt(109)#

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

In this case we have
#color(white)("XXX")x^2+(y-6)^2=109#
graph{x^2+(y-6)^2=109 [-14.24, 14.23, -7.12, 7.11]}

Feb 15, 2016

Answer:

#x^2+y^2+8x+6y-72=0#

Explanation:

It means that #(-4,-3)# is center and radius is distance between #(-4,-3)# and #(0,6)#. The radius is hence given by

#sqrt{(0-(-4))^2+(6-(-3))^2)# or #sqrt(16+81)# or #sqrt87#

Hence equation of circle is

#(x-(-4))^2+(y-(-3^2))=87# or

#(x+4)^2+(y+3)^2=87#

#x^2+8x+16+y^2+6y+9=87# or

#x^2+y^2+8x+6y+16+9-87=0# or

#x^2+y^2+8x+6y-72=0#