ExamBro
ExamBro
MHT CET · Maths · Circle

Number of common tangents to the circles \(x^2+y^2-6 x-14 y+48=0\) and \(x^2+y^2-6 x=0\) are

  1. A \(0\)
  2. B \(1\)
  3. C \(4\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

\( x^2+y^2-6 x-14 y+48=0 \)
\( \therefore \mathrm{C}_1(3,7), \mathrm{r}_1=\sqrt{10} \)
\( \text {Again } x^2+y^2-6 x=0 \)
\( \therefore \mathrm{C}_2(3,0), \mathrm{r}_2=3\)\
Now \(l\left(\mathrm{C}_1 \mathrm{C}_2\right)=\) distance between centres
\(\therefore l\left(\mathrm{C}_1 \mathrm{C}_2\right)=\sqrt{0^2+7^2}=7 \text { and } \)
\( \mathrm{r}_1+\mathrm{r}_2=\sqrt{10}+3 < l\left(\mathrm{C}_1 \mathrm{C}_2\right)\)
\(\Rightarrow\) The given circles are disjoint.
\(\Rightarrow\) Number of common tangents is 4 .