Question

Equation for a circle with a diameter that has endpoints at (2, –5) and (8, –9)

Answer 1

The standard form for the equation of a circle is :

 (x−h)^2+(y−k)^2=r2 ----------- EQ(1)
 where handk are the x and y coordinates of the center of the circle and r is the radius.
 The center of the circle is the midpoint of the diameter.

 So the midpoint of the diameter with endpoints at (2,-5)and(8,-9) is :

 ((2+(8))/2,(-5+(-9))/2)=(5,-7)

 So the point (5,-7) is the center of the circle.

  Now, use the distance formula to find the radius of the circle:

  r^2=(2−(5))^2+(-5−(-7))^2=9+4=13

 ⇒r=√13

 Subtituting h=5, k=-7 and r=√13 into EQ(1) gives :

 (x-5)^2+(y+7)^2=13