Find the equation of the circle which passes through the origin and the point of intersection of the circles x^2+y^2-4x-8y+16=0 and x^2+y^2+6x-4y-3=0?

1 Answer
Apr 8, 2018

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The equation of any circle which passes through the point of intersection of the circles

#x^2+y^2-4x-8y+16=0 color(red)" circle=1"#

and #x^2+y^2+6x-4y-3=0color(blue)" circle=2"#

may be written as

#lambda(x^2+y^2-4x-8y+16) + x^2+y^2+6x-4y-3=0#,where #lambda # is a parameter.

This circle also passes through the origin.

So #lambda(0^2+0^2-4*0-8*0+16) + 0^2+0^2+6*0-4*0-3=0#

#=>lambda=3/16#

Hence required equation of th circle

#3/16(x^2+y^2-4x-8y+16 + x^2+y^2+6x-4y-3=0#

#=>3x^2+3y^2-12x-24y+48 +16x^2+16y^2+96x-64y-48=0#

#=>19x^2+19y^2+84x-88y=0color(green)" circle=3"#