JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
The number of common tangents to the circles \({x^2} + {y^2} - 4x - 6y - 12 = 0\) and \({x^2} + {y^2} + 6x + 18y + 26 = 0\) is
- A \(4\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(c_{1}(2,3);\) \(r_{1}=\sqrt{4+9+12}=5\) And \(c_{2}(-3,-9)\) \(r_{2}=\sqrt{9+81-26}=8\) \(\therefore \quad\) Distance \(c_{1} c_{2}=\sqrt{25+144}=13\) \(\therefore \quad c_{1} c_{2}=r_{1}+r_{2}\) touching externally. \(\Rightarrow \quad 3\) common tangents.
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