Question

If two 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. $f_{1}g=f{g}_1$
• B. $f{f}_1=g{g}_1$
• C. $f^2+g^2=f_1^2+g_1^2$
• D. None of these

Answer 1

The correct option is A $f_{1}g=f{g}_1$

$x^2+y^2+2gx+2fy=0$ has center as $(-g,-f)$ and radius as $\sqrt{g^2+f^2}$

$x^2+y^2+2g_{1}x+2f_{1}y=0$ has center as $(-g_1,-f_1)$ and radius as $\sqrt{g_1^2+f_1^2}$

Since the two circles touch each other, distance between centers equals sum or difference between radii.

$\sqrt{(f-f_1{)}^2+(g-g_1{)}^2}=\sqrt{g^2+f^2}\pm \sqrt{g_1^2+f_1^2}$

Squaring both sides, $(f-f_1{)}^2+(g-g_1{)}^2=g^2+f^2+g_1^2+f_1^2\pm 2\sqrt{(g^2+f^2)(g_1^2+f_1^2)}$

$\Rightarrow -2f{f}_1-2g{g}_1=\pm 2\sqrt{(g^2+g_1^2)(f^2+f_1^2)}$

Squaring both sides, we get $f^2{f}_1^2+g^2{g}_1^2+2f{f}_{1}g{g}_1=f^2{g}^2+f^2{g}_1^2+f_1^2{g}^2+f_1^2{g}_1^2$

Dividing throughout by $f{f}_1.g{g}_1$, we get

$\frac{f{f}_1}{g{g}_1}+\frac{g{g}_1}{f{f}_1}+2=\frac{fg}{f_1{g}_1}+\frac{f{g}_1}{f_{1}g}+\frac{f_{1}g}{f{g}_1}+\frac{f_1{g}_1}{fg}$

These two sides become equal when $f_{1}g=f{g}_1$