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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If a circle \(C\) passing through \((4, 0)\) touches the circle \(x^2 + y^2 + 4x - 6y - 12 = 0\) externally at a point \((1, -1),\) then the radius of the circle \(C\) is
- A \(5\)
- B \(2\sqrt 5\)
- C \(4\)
- D \(\sqrt {57}\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
Let \(A\) be the center og given circle and \(B\) be the center of circle \(C\). \({x^2} + {y^2} + 4x - 6y - 1 = 0\) \(\therefore A = \left( { - 2,3} \right)\) and \(B = \left( {g,f} \right)\) Now, from the figure, we have \(\frac{{ - 2 + g}}{2} = 1\,\) and…
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