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The equation of any circle passing through the origin and with its centre on the $X$-axis is given by

  1. $x^2+y^2-2ax=0$ where $a$ must be positive
  2. $x^2+y^2-2ax=0$ for any given $a \in \mathbb{R}$
  3. $x^2+y^2-2by=0$ where $b$ must be positive
  4. $x^2+y^2-2by=0$ for any given $b \in \mathbb{R}$
in Numerical Ability by Veteran (425k points)
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1 Answer

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Answer: $\mathbf A$

If a circle passes through the origin and the centre lies on the $\mathrm x$ axis then the length of the abscissa will be equal to the circle's radius and the ordinate(y-coordinate) will be $0$.

So, the equation of the circle will be of the form:

$\mathrm {(x-a)^2 + y^2 = a^2}$

$\Rightarrow \mathrm{ x^2 + y^2 -2ax = 0}$

$\therefore \mathbf A $ is the right option.
by Boss (14k points)
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@ankitgupta.1729

Can you please checkout this.

I have doubt between options $a$ and $b$.

0

@`JEET

'a' can be negative, right ? Coordinates of centre can be  (3,0) or (-3,0). Then equation will be $x^2 + y^2 + 6x =0 $ (or) $x^2 + y^2 - 6x =0 $ and In both cases, it will satisfy the given condition. right ?

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