How to find the standard form of the equation of the specified circle given that it passes through the origin and has its center at (-5, 4)?

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Answer 1 · 2018-06-05T08:14:29

#x^2+y^2+10x-8y=0#

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

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

we need to find the radius

we can use Pythagoras

since the circle passes through the origin we have

#r^2=(0--5)^2+(0-4)^2#

#r^2=25+16=41#

#:. " eqn "(x- -5)^2+(y-4)^2=41#

#(x+5)^2+(y-4)^2=41#

#x^2+10x+25+y^2-8y+16=41#

#=>x^2+y^2+10x-8y=0#