Find equation of circle with center on a line y= - x has radius 4 and passes through origin?

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Answer 1 · 2018-03-05T16:12:48

Equation of circle is #x^2+y^2+4sqrt2x-4sqrt2y=0#

As circle passes thrugh origin, its cinstant term ought to be #0# and hence equation is of the type

#x^2+y^2+2gx+2fy=0#, whose center is #(-g,-f)# and radius is #sqrt(g^2+f^2)#

As center #(-g,-f)# lies on #y=-x#,

we have #-f=-(-g)# or #f=-g#

and aas radius is #4#, we have #sqrt(g^2+f^2)=4#

or #2g^2=16# i.e. #g=sqrt8=2sqrt2# and #f=-2sqrt2#

and equation of circle is #x^2+y^2+4sqrt2x-4sqrt2y=0#

graph{(x^2+y^2+4sqrt2x-4sqrt2y)(y+x)=0 [-12.79, 7.21, -2.92, 7.08]}