How do you write an equation for a circle whose diameter has endpoints (-2, 3) and (4, -1)?

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Jim G. Share
Jul 16, 2016

Answer:

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

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 coordinates of the centre and r, the radius.

To establish the equation, we require to know it's centre and radius.

Since we are given the endpoints of the diameter. Then the centre will be at the midpoint and the radius will be the distance from the centre to either of the 2 endpoints.

The midpoint can be calculated using the #color(blue)"midpoint formula"#

#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)|)))#
where # (x_1,y_1)" and " (x_2,y_2)" are 2 points"#

The 2 points here are (-2 ,3) and (4 ,-1)

#rArr " centre" =(1/2(-2+4),1/2(3-1))=(1,1)#

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

#color(red)(|bar(ul(color(white)(a/a)color(black)(r=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 the centre (1 ,1) and the endpoint (-2 ,3)

#r=sqrt((-2-1)^2+(3-1)^2)=sqrt(9+4)=sqrt13#

The equation of the circle can now be written.

#(x-1)^2+(y-1)^2=(sqrt13)^2#

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

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