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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

  1. A 4
  2. B 2
  3. C 3
  4. D 1
Verified Solution

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 \(=3\)