KCET · Maths · Circle
The total number of common tangents of \(x^{2}+y^{2}-6 x-8 y+9=0\) and \(x^{2}+y^{2}=1\) is
- A 4
- B 2
- C 3
- D 1
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
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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