Respuesta :
Answer:
x^2+y^2-2x-8y-8=0
Explanation:
Let, (-3,1) => (x1,y1) and (5,7) => (x2,y2) are endpoints of diameter of a circle
We know that if (x1,y1) and (x2,y2) are ends of the diameter then equation of circle is given
by. (x - x1)(x - x2) + (y - y1)(y - y2) =0
So (x-(-3))(x - 5) + (y - 1)(y - 7)=0
(x+3)(x-5)+(y-1)(y-7)=0
x^2+3x-5x-15+y^2-y-7y+7=0
x^2+y^2-2x-8y-8=0 is the required equation of circle.
x^2+y^2-2x-8y-8=0
Explanation:
Let, (-3,1) => (x1,y1) and (5,7) => (x2,y2) are endpoints of diameter of a circle
We know that if (x1,y1) and (x2,y2) are ends of the diameter then equation of circle is given
by. (x - x1)(x - x2) + (y - y1)(y - y2) =0
So (x-(-3))(x - 5) + (y - 1)(y - 7)=0
(x+3)(x-5)+(y-1)(y-7)=0
x^2+3x-5x-15+y^2-y-7y+7=0
x^2+y^2-2x-8y-8=0 is the required equation of circle.
