What is the equation of the circle with endpoints of the diameter of a circle are (7,4) and (-9,6)?

1 Answer
May 25, 2016

Answer:

#(x+1)^2+(y-5)^2=65#

Explanation:

The standard form of the equation of a circle is.

#color(red)(|bar(ul(color(white)(a/a)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(a/a)|)))#
where (a ,b) are the coords of the centre and r ,the radius.

We require to know the centre and radius to establish the equation.

Given the coords of the endpoints of the diameter , then the centre of the circle will be at the mid-point.

Given 2 points #(x_1,y_1)" and " (x_2,y_2)# then the mid-point is.

#color(red)(|bar(ul(color(white)(a/a)color(black)(1/2(x_1+x_2),1/2(y_1+y_2))color(white)(a/a)|)))#

The mid-point of (7 ,4) and (-9 ,6) is therefore.

#=(1/2(7-9),1/2(4+6))=(-1,5)=" centre"#

Now the radius is the distance from the centre to either of the 2 endpoints.

Using the #color(blue)"distance formula"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(a/a)|)))#
where #(x_1,y_1)" and " (x_2,y_2)" are 2 points"#

The 2 points here are centre (-1 ,5) and endpoint (7 ,4)

#d=sqrt((-1-7)^2+(5-4)^2)=sqrt65=" radius"#

We now have centre = (a ,b) = (-1 ,5) and r #=sqrt65#

#rArr(x+1)^2+(y-5)^2=65" is equation of circle"#