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

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

The academic affairs council is the highest expert body processing academic affairs, as stipulated by the university statutes. specialist aaro häkkinen (educational policy) took part in the council’s meeting. the agenda of the meeting included bachelors’ feedback, the timing of pedagogical studies, the grant system for tuition fees, the standardisation of administrative practices for internships and the situation of foreign-language and multilingual master’s programmes.