Give the equation of the circle whose center is (5,-3) and goes through (2,5)?

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Answer 1 · 2018-05-03T20:43:51

#(x -5)^2 + (y--3)^2=(sqrt73)^2#

The standard Cartesian form for the equation of a circle is:

#(x -h)^2 + (y-k)^2=r^2" [1]"#

where #(x,y)# is any point on the circle, #(h,k)# is the center point, and #r# is the radius.

Substitute #(h,k) = (5,-3)# into equation [1]:

#(x -5)^2 + (y--3)^2=r^2" [1.1]"#

To find the value of #r#, substitute #(x,y) = (2,5) into equation [1.1]:

#(2 -5)^2 + (5--3)^2=r^2#

#r^2 = 73#

#r = sqrt73#

Substitute #r = sqrt73# into equation [1.1]:

#(x -5)^2 + (y--3)^2=(sqrt73)^2#