Question

Match the following
Circle are orthogonal

Answer 1

$x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g^{\prime }x+2f^{\prime }y+c^{\prime }=0$
are orthogonal when $2g{g}^{\prime }+2f{f}^{\prime }-(c+c^{\prime })=0$
A) For $x^2+y^2+4x+5=0$ and $x^2+y^2+6y-5=0$
$2g{g}^{\prime }+2f{f}^{\prime }-(c+c^{\prime })=\mathrm{2.2.0}+\mathrm{2.0.3}-(5-5)=0$
B) For $x^2+y^2+6y-7=0$ and $x^2+y^2+4x+7=0$
$2g{g}^{\prime }+2f{f}^{\prime }-(c+c^{\prime })=\mathrm{2.0.2}+\mathrm{2.3.0}-(-7+7)=0$
C) For $x^2+y^2+4x+6y+6=0$ and $x^2+y^2+2x+4y+10=0$
$2g{g}^{\prime }+2f{f}^{\prime }-(c+c^{\prime })=\mathrm{2.2.1}+\mathrm{2.3.2}-(6+10)=0$
And
For $x^2+y^2+4x+6y+6=0$ and $x^2+y^2-2x+4y+2=0$
$2g{g}^{\prime }+2f{f}^{\prime }-(c+c^{\prime })=\mathrm{2.2.}(-1)+\mathrm{2.3.2}-(6+2)=0$
D) For $x^2+y^2+6x-2y-8=0$ and $x^2+y^2+2x+4y+10=0$
$2g{g}^{\prime }+2f{f}^{\prime }-(c+c^{\prime })=\mathrm{2.3.1}+2.(-1).2-(-8+10)=0$