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

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

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