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The total number of common tangents of $x^{2}+y^{2}-6 x-8 y+9=0$ and $x^{2}+y^{2}=1$ is
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Verified Answer
The correct answer is:
3
The given equation of circles
and
$$
\begin{gathered}
x^{2}+y^{2}-6 x-8 y+9=0 \\
x^{2}+y^{2}=1
\end{gathered}
$$
Radius and centre of both circles is
$C_{1} \rightarrow(3,4), R_{1}=\sqrt{9+16-9}=4$ $C_{2} \rightarrow(0,0), R_{2}=1$ $C_{1} C_{2}=\sqrt{(3-0)^{2}+(4-0)^{2}}=\sqrt{9+16}=5$ $$ R_{1}+R_{2}=4+1=5 $$ $$ C_{1} C_{2}=R_{1}+R_{2} $$ distinct while the transverse tangents are coincident.
So, number of comman tangents $=3$
and
$$
\begin{gathered}
x^{2}+y^{2}-6 x-8 y+9=0 \\
x^{2}+y^{2}=1
\end{gathered}
$$
Radius and centre of both circles is
$C_{1} \rightarrow(3,4), R_{1}=\sqrt{9+16-9}=4$ $C_{2} \rightarrow(0,0), R_{2}=1$ $C_{1} C_{2}=\sqrt{(3-0)^{2}+(4-0)^{2}}=\sqrt{9+16}=5$ $$ R_{1}+R_{2}=4+1=5 $$ $$ C_{1} C_{2}=R_{1}+R_{2} $$ distinct while the transverse tangents are coincident.
So, number of comman tangents $=3$
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