Question

If the circles $x^2+y^2+2gx+2fy=0$, and $x^2+y^2+2g^{1}x+2f^{1}y=0$ touch each other, then

• A. $fg=f^1{g}^1$
• B. $f^{1}g=f{g}^1$
• C. $f{f}^1=g{g}^1$
• D. None of these

Answer 1

Answer: (B)

$f^{1}g=f{g}^1$

If $C_1$ and $C_2$ are the centres of the circles respectively and R and r are their corresponding radii then if the circles touch each other,
$C_1{C}_2=R+r$
Consider equation 1
$(x+g{)}^2+(y+f{)}^2=g^2+f^2$
Similarly,

$(x+g^1{)}^2+(y+f^1{)}^2=\left(g^1\right){}^2+\left(f^1\right){}^2$

Therefore

$C^1{C}_2=R+r$ implies,

$\sqrt{(g-g^1{)}^2+(f-f^1{)}^2}=\sqrt{g^2+f^2}+\sqrt{\left(g^1\right){}^2+\left(f^1\right){}^2}$

Squaring both sides
$(g-g^1{)}^2+(f-f^1{)}^2=g^2+(g^1{)}^2+f^2+(f^1{)}^2+2\sqrt{g^2+f^2}.\sqrt{\left(g^1\right){}^2+\left(f^1\right){}^2}$
$g^2+(g^1{)}^2+f^2+(f^1{)}^2-2g{g}^1-2f{f}^1=g^2+(g^1{)}^2+f^2+(f^1{)}^2+2\sqrt{g^2+f^2}.\sqrt{\left(g^1\right){}^2+\left(f^1\right){}^2}$
$-2g{g}^1-2f{f}^1=2\sqrt{g^2+f^2}.\sqrt{\left(g^1\right){}^2+\left(f^1\right){}^2}$
$-g{g}^1-f{f}^1=\sqrt{g^2+f^2}.\sqrt{\left(g^1\right){}^2+\left(f^1\right){}^2}$

Squaring both sides, we get
$(g{g}^1{)}^2+(f{f}^1{)}^2+2g{g}^{1}f{f}^1=(g^2+f^2).\left(\right(g^1{)}^2+\left(f^1{)}^2\right)$

$(g{g}^1{)}^2+(f{f}^1{)}^2+2g{g}^{1}f{f}^1=(g{g}^1{)}^2+(f{f}^1{)}^2+(g{f}^1{)}^2+(f{g}^1{)}^2$

$2g{g}^{1}f{f}^1=(g{f}^1{)}^2+(f{g}^1{)}^2$

$(g{f}^1{)}^2+(f{g}^1{)}^2-2g{g}^{1}f{f}^1=0$

$(g{f}^1-f{g}^1{)}^2=0$

$g{f}^1=f{g}^1$