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)?

1 Answer
Jun 5, 2018

x^2+y^2+10x-8y=0x2+y2+10x8y=0

Explanation:

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

(x-a)^2+(y-b)^2=r^2(xa)2+(yb)2=r2

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)^2r2=(05)2+(04)2

r^2=25+16=41r2=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