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
- A \(0\)
- B \(1\)
- C \(4\)
- D \(2\)
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 .
\( \therefore \mathrm{C}_1(3,7), \mathrm{r}_1=\sqrt{10} \)
\( \text {Again } x^2+y^2-6 x=0 \)
\( \therefore \mathrm{C}_2(3,0), \mathrm{r}_2=3\)\
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 .
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