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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If the circle \(x^2 + y^2 - 6x - 8y + (25 - a^2)\, = 0\) touches the axis of \(x\), then \(a\) equals
- A \(0\)
- B \( \pm 4\)
- C \( \pm 2\)
- D \( \pm 3\)
Answer & Solution
Correct Answer
(B) \( \pm 4\)
Step-by-step Solution
Detailed explanation
\({x^2} + {y^2} - 6x - 8y + \left( {25 - {a^2}} \right) = 0\) Radius \( = 4 = \sqrt {9 + 16 + \left( {25 - {a^2}} \right)} \) \( \Rightarrow a = \pm 4\)
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