How would you find the equation of the circle, center at (5,-3) and radius of 4?

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Answer 1 · 2018-03-04T10:18:56

#x^2+y^2-10x+6y+18=0#

the eqn of a circle with centre #(a,b)# radius #r#

is given by

#(x-a)^2+(y-b)^2=r^2#

in this case we have

centre #(5,-3)#, radius#" "r=4#

eqn

#(x- 5)^2+(y- -3)^2=4^2#

#=>(x- 5)^2+(y+3)^2=4^2#

expanding

#x^2-10x+25+y^2+6y+9=16#

#x^2+y^2-10x+6y+18=0#