If h, k, p, q neq 0 and the circles x^2+y^2+2 | vedclass

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(A) The given circles are $x^2+y^2+2hx+2ky=0$ and $x^2+y^2+2px+2qy=0$. Both circles pass through the origin $(0,0)$. Two circles $x^2+y^2+2g_1x+2f_1y=

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$-1$

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(A) The given circles are $x^2+y^2+2hx+2ky=0$ and $x^2+y^2+2px+2qy=0$. Both circles pass through the origin $(0,0)$. Two circles $x^2+y^2+2g_1x+2f_1y=0$ and $x^2+y^2+2g_2x+2f_2y=0$ touch each other at the origin if their centers are collinear with the origin,which implies $\frac{g_1}{f_1} = \frac{g_2}{f_2}$,or $g_1f_2 = g_2f_1$. Here,$g_1=h, f_1=k, g_2=p, f_2=q$. Thus,the condition for touching at the origin is $hq = pk$,which implies $hq - pk = 0$. Also,since $hq = pk$ and $p, k 
eq 0$,we have $\frac{hq}{pk} = 1$. Substituting these values into the expression: $hq - pk - \frac{hq}{pk} = 0 - 1 = -1$.

If $h, k, p, q \neq 0$ and the circles $x^2+y^2+2hx+2ky=0$ and $x^2+y^2+2px+2qy=0$ touch each other at the origin,then $hq-pk-\frac{hq}{pk}$ is equal to

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