If the circles x^2+y^2+2hx+2ky=0 and x^2+y^2+ | vedclass

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(B) The given circles are $C_1: x^2+y^2+2hx+2ky=0$ and $C_2: x^2+y^2+2h'x+2k'y=0$. The centers are $O_1 = (-h, -k)$ and $O_2 = (-h', -k')$. The radii

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(B) The given circles are $C_1: x^2+y^2+2hx+2ky=0$ and $C_2: x^2+y^2+2h'x+2k'y=0$. The centers are $O_1 = (-h, -k)$ and $O_2 = (-h', -k')$. The radii are $r_1 = \sqrt{h^2+k^2}$ and $r_2 = \sqrt{h'^2+k'^2}$. Since both circles pass through the origin $(0,0)$,they touch each other if and only if the distance between their centers is equal to the sum or difference of their radii. The distance between centers is $d = \sqrt{(-h+h')^2 + (-k+k')^2} = \sqrt{(h-h')^2 + (k-k')^2}$. The condition for touching is $d^2 = (r_1 \pm r_2)^2 = r_1^2 + r_2^2 \pm 2r_1r_2$. Substituting the values: $(h-h')^2 + (k-k')^2 = (h^2+k^2) + (h'^2+k'^2) \pm 2\sqrt{h^2+k^2}\sqrt{h'^2+k'^2}$. $h^2 - 2hh' + h'^2 + k^2 - 2kk' + k'^2 = h^2 + k^2 + h'^2 + k'^2 \pm 2\sqrt{h^2+k^2}\sqrt{h'^2+k'^2}$. $-2hh' - 2kk' = \pm 2\sqrt{h^2+k^2}\sqrt{h'^2+k'^2}$. $-(hh' + kk') = \pm \sqrt{h^2+k^2}\sqrt{h'^2+k'^2}$. Squaring both sides: $(hh' + kk')^2 = (h^2+k^2)(h'^2+k'^2)$. $h^2h'^2 + k^2k'^2 + 2hh'kk' = h^2h'^2 + h^2k'^2 + k^2h'^2 + k^2k'^2$. $2hh'kk' = h^2k'^2 + k^2h'^2$. $h^2k'^2 - 2hh'kk' + k^2h'^2 = 0$. $(hk' - kh')^2 = 0$. Therefore,$hk' = kh'$,which implies $\frac{h'k}{hk'} = 1$.

If the circles $x^2+y^2+2hx+2ky=0$ and $x^2+y^2+2h'x+2k'y=0$ touch each other,then $\frac{h'k}{hk'} = $

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