How do we find the standard equation of the circle with endpoints of a diameter #(1/2, 4)# #(3/2, -1)#?

2 Answers
Jun 15, 2017

Answer:

#(x-1)^2+(y-3/2)^2=2.55^2#

Explanation:

The standard equation of a circle takes the form:

#(x-a)^2+(y-b)^2=c^2# where #c# is the radius

We need to find the radius, #r#, and the location of the centre, the point #(a,b)#

The length of the diameter is given by:

#d = 2r = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)#

#d = sqrt((3/2-1/2)^2 + (-1-4)^2)#
#= sqrt(1^2+(-5)^2) = sqrt(1+25) = 5.1# units

The radius will be half the diameter, #2.55# units

We can then use the midpoint formula to find the coordinates of the centre. The midpoint formula simply averages the values of the endpoints:

#(a,b) = ((x_1+x_2)/2,(y_1+y_2)/2) = ((1/2+3/2)/2,(4-1)/2)=(1,3/2)#

Now we have all the information we need to write the equation for the circle:

#(x-1)^2+(y-3/2)^2=2.55^2#

Jun 15, 2017

Answer:

#(x-1)^2+(y-3/2)^2=13/2#

Explanation:

#"the standard form of the equation of a circle 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 the radius.

#"the centre will be at the midpoint of the diameter"#

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

#rArr"centre " =(1,3/2)larr(a,b)#

#"the radius is the distance from the centre to either of "#
#"the endpoints"#

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

#"points are " (1,3/2)" and " (3/2,-1)#

#r=sqrt((3/2-1)^2+(-1-3/2)^2)#

#color(white)(r)=sqrt(1/4+25/4)=sqrt(13/2)#

#rArr(x-1)^2+(y-3/2)^2=(sqrt(13/2))^2#

#rArr(x-1)^2+(y-3/2)^2=13/2" is the equation"#