centre of the circle =Midpoint of the diameter
centre of the circle=
[tex]( \frac{ - 8 + 4}{2} , \frac{5 + 1}{2} ) = ( - 2 ,3)[/tex]
radius of the circle=
[tex]\sqrt{(4 - ( - 2) ^{2} + (5 - 3 )^{2} } = \sqrt{36 + 4} = \sqrt{40 } = 2 \sqrt{10}
[/tex]
equation of the circle: (x+2)²+(y-3)² = 40
ie.. x²+4x+4+ y²- 6y+ 9=40
ie... x²+y²+4x-6y-27=0
this is the answer......
