How do you write the equation of the circle with diameter that has endpoints at (–1, –6) and (–3, –2).?

1 Answer
Jun 30, 2016

Answer:

#x^2+y^2+4x+8y=0#

Explanation:

The center of the circle is the middle point of the diameter.

Then its coordinates are obtained by using

#((x_1+x_2)/2,(y_1+y_2)/2)#

in this case:

#((-1-3)/2;(-6-2)/2)#

#(-2;-4)#

The radius is the half of the diameter, that is obtained by calculating the distance

#sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

in this case:

#sqrt((-3+1)^2+(-2+6)^2)#

#sqrt(4+16)=sqrt(20)#

To obtain teh equation of the circle you can substitute in the general equation #x_0=-2, y_0=-4, r=sqrt(20)#:

(the general equation is #(x-x_0)^2+(y-y_0)^2=r^2#)

#(x+2)^2+(y+4)^2=20#

#x^2+4+4x+y^2+16+8y-20=0#

#x^2+y^2+4x+8y=0#