Question

If the circle $x^2+y^2+2gx+2fy=0$ and $x^2+y^2+2g^{\prime }x+2f^{\prime }y=0$ touch each other then

• A. $f{f}^{\prime }=g{g}^{\prime }$
• B. $fg=f^{\prime }{g}^{\prime }$
• C. $(fg{)}^2=(f^{\prime }{g}^{\prime }{)}^2$
• D. $f{g}^{\prime }=f^{\prime }g$

Answer 1

Answer: (D)

$f{g}^{\prime }=f^{\prime }g$

When two circles touch each other, distance between their centres equals either the sum of their radii or the difference between their radii.

$\therefore \sqrt{(g-g^{\prime }{)}^2+(f-f^{\prime }{)}^2}=\sqrt{g^2+f^2}\pm \sqrt{g^{\prime 2}+f^{\prime 2}}$

Squaring both sides, we get

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

$\therefore -2g{g}^{\prime }-2f{f}^{\prime }=\pm 2\sqrt{(g^2+f^2)(g^{\prime 2}+f^{\prime 2})}$

Squaring both sides, we get $g^2{g}^{\prime 2}+f^2{f}^{\prime 2}+2g{g}^{\prime }f{f}^{\prime }=g^2{g}^{\prime 2}+g^2{f}^{\prime 2}+f^2{g}^{\prime 2}+f^2{f}^{\prime 2}$

$\Rightarrow 2g{g}^{\prime }f{f}^{\prime }=g^2{f}^{\prime 2}+g^{\prime 2}{f}^2$

$\therefore g^2{f}^{\prime 2}-2g{g}^{\prime }f{f}^{\prime }+g^{\prime 2}{f}^2=0$

$\therefore g{f}^{\prime }=g^{\prime }f$