Write the standard equation of the circle which has (-1,3) and (2,4) as the endpoints of a diameter?

1 Answer
Jun 16, 2018

#(x-1/2)^2+(y-7/2)^2=5/2#

Explanation:

#"the equation of a circle in standard form is"#

#color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))#

#"where "(a,b)" are the coordinates of the centre and r"#
#"is the radius"#

#"given the endpoints of the diameter then the centre of"#
#"the circle is at the midpoint and the radius is the distance"#
#"from the centre to either of the endpoints"#

#"the midpoint is the average of the coordinates of "#
#"the endpoints"#

#"midpoint "=[1/2(-1+2),1/2(3+4)]#

#color(white)(xxxxxxx)=(1/2,7/2)larrcolor(red)"centre"#

#"to calculate the radius use the "color(blue)"distance formula"#

#•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#"let "(x_1,y_1)=(1/2,7/2)" and "(x_2,y_2)=(2,4)#

#r=sqrt((2-1/2)^2+(4-7/2)^2)#

#color(white)(r)=sqrt(9/4+1/4)=sqrt(10/4)#

#"the equation of the circle is therefore"#

#(x-1/2)^2+(y-7/2)^2=(sqrt(10/4))^2#

#(x-1/2)^2+(y-7/2)^2=5/2#