KCET · Maths · Circle
The number of common tangents to the circles \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}-6 x-8 y-24=0\) is
- A 3
- B 4
- C 2
- D 1
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
Given, equation of circles are
\[
x^{2}+y^{2}=4
\]
whose radius \(=2\)
centre \(=(0,0)\)
and \(x^{2}+y^{2}-6 x-8 y-24=0\),
whose radius \(=\sqrt{9+16-24}=\sqrt{1}=1\)
and centre \(=(3,4)\)
Now, \(\quad c_{1} c_{2}=\sqrt{(3-0)^{2}+(4-0)^{2}}\)
\[
=\sqrt{9+16}=5
\]
\[
a_{1}+a_{2}=2+1=3
\]
Since, \(c_{1} c_{2}>a_{1}+a_{2}\)
\(\therefore\) Number of common tangents \(=4\)
\[
x^{2}+y^{2}=4
\]
whose radius \(=2\)
centre \(=(0,0)\)
and \(x^{2}+y^{2}-6 x-8 y-24=0\),
whose radius \(=\sqrt{9+16-24}=\sqrt{1}=1\)
and centre \(=(3,4)\)
Now, \(\quad c_{1} c_{2}=\sqrt{(3-0)^{2}+(4-0)^{2}}\)
\[
=\sqrt{9+16}=5
\]
\[
a_{1}+a_{2}=2+1=3
\]
Since, \(c_{1} c_{2}>a_{1}+a_{2}\)
\(\therefore\) Number of common tangents \(=4\)
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