How do you write the standard form of the equation of the circle with the given the diameter with endpoints (-10, -6) and (-2, -4)?

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Answer 1 · 2016-11-06T02:37:29

#(x+6)^2 +(y+5)^2 = 17#

Diameter with endpoints #(-10,-6)# and #(-2,-4)#

To find the length of the diameter, use the distance formula

#d=sqrt((x_1-x_2)^2+(y_1-y_2)^2)#
where #d# is the distance between two points #(x_1,y_1)# and #(x_2,y_2)#

#d=sqrt((-10 - -2)^2 +(-6 - -4)^2)#

#d=sqrt((-8)^2+(-2)^2)#

#d=sqrt(64+4)#

#d=sqrt68=2sqrt17#

The length of the radius #r# equals the diameter divided by 2.

#r=(2sqrt17)/2=sqrt17#

The center of the circle is the midpoint of the diameter.

Remember, the midpoint of a segment is just the average of the x coordinates and the average of the y coordinates. This gives the midpoint formula

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

midpoint #=((-10+ -2)/2, (-6 + -4)/2)= (-6,-5)#

The center is #(-6,-5)# and the radius is #sqrt17#

The equation of a circle is #(x-h)^2 +(y-k)^2 =r^2#,
where #(h,k)# is the center and #r# is the radius.

Therefore, the equation is

#(x- -6)^2 + (y- -5)^2 = (sqrt17)^2#

#(x+6)^2 +(y+5)^2 = 17#