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प्रश्न
If the circle x2 + y2 + 2ax + 8y + 16 = 0 touches x-axis, then the value of a is
विकल्प
± 16
±4
± 8
±1
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उत्तर
± 4
The equation of the circle is x2 + y2 + 2ax + 8y + 16 = 0.
Its centre is \[\left( - a, - 4 \right)\] and its radius is a units.
Since the circle touches the x-axis, we have:
\[\sqrt{\left( - a + a \right)^2 + \left( 4 - 0 \right)^2} = a\]
⇒ \[a = \pm 4\]
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