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If $h k p q \neq 0$ and the circles $x^2+y^2+2 h x+2 k y=0$ and $x^2+y^2+2 p x+2 q y=0$ touch each other at the origin, then $h q-p k-\frac{h q}{p k}$ is equal to
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Verified Answer
The correct answer is:
−1
As we know that two circles
$$
\begin{aligned}
& x^2+y^2+2 g x+2 f y=0 \text { and } \\
& x^2+y^2+2 g^{\prime} x+2 f^{\prime} y=0 \text { touch each other } \\
& \text { if } g f^{\prime}=g^{\prime} f
\end{aligned}
$$
So, circles given in question, touch each other if
$$
\begin{aligned}
h q-p k=0 \text { and } \frac{h q}{p k}=1 \\
h q=p k \\
\therefore \quad h q-p k-\frac{h q}{p k}=0-1=-1
\end{aligned}
$$
$$
\begin{aligned}
& x^2+y^2+2 g x+2 f y=0 \text { and } \\
& x^2+y^2+2 g^{\prime} x+2 f^{\prime} y=0 \text { touch each other } \\
& \text { if } g f^{\prime}=g^{\prime} f
\end{aligned}
$$
So, circles given in question, touch each other if
$$
\begin{aligned}
h q-p k=0 \text { and } \frac{h q}{p k}=1 \\
h q=p k \\
\therefore \quad h q-p k-\frac{h q}{p k}=0-1=-1
\end{aligned}
$$
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