Question

Find the standard form of the equation of the circle with endpoints of a diameter of the points (1,8) and (-9,6) ... Type the standard form of the equation of this circle....

Answer 1

the standard form of a circle is x^2 + y^2 + 2gx + 2fy +c =0
here (-g,-f) is center and sqrt(g^2 + f^2 -c) is radius..
if (x1,y1) and (x2,y2) are end point of circle,,the eqn is
(x-x1)(x-x2) + (y-y1)(y-y2) = 0
i wont type the final ans though,,you should do yourself now..

Answer 2

thank you

Answer 3

the midpoint of the segment will be the centre. half the length will be the radius

midpoint x = (1 - 9)/2 y = (8 + 6)/2

midpoint (-4, 7)

length = (-8)^2 + (14)^2

radius = \[radius = \sqrt{130}\]

equation of the circle

(x +4)^2 + (y -7)^2 = 130

Answer 4

thank you

Answer 5

@ my previous point (x1,y1) and (x2,y2) are end points of diameter of circle **

Answer 6

:-) i understood

Answer 7

oops radius should be \[r = \sqrt{57}\]