JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
The common tangent to the circles \(x^2 + y^2 = 4\) and \(x^2 + y^2 + 6x + 8y - 24 = 0\) also passes through the point
- A \((-4, 6)\)
- B \((6, -2)\)
- C \((-6, 4)\)
- D \((4, -2)\)
Answer & Solution
Correct Answer
(B) \((6, -2)\)
Step-by-step Solution
Detailed explanation
Circle \({x^2} + {y^2} = 4\) \( \Rightarrow {c_1}\left( {0,0} \right);{r_1} = 2\) and circle \({x^2} + {y^2} + 6x + 8y - 24 = 0\) \( \Rightarrow {c_2}\left( { - 3,4} \right);{r_2} = 7\) \( \Rightarrow d = {c_1}{c_2} = 5\) also \(d = \left| {{r_1} - {r_2}} \right|\) circles touch…
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