If the circle $x^2+y^2-6 x-8 y+\left(25-a^2\right)=0$ touches the axis of $x$, then a equals.

If the circle $x^2+y^2-6 x-8 y+\left(25-a^2\right)=0$ touches the axis of $x$, then a equals.
  1. 0
  2. $\pm 4$
  3. $\pm 2$
  4. $\pm 3$

Solution


$ \begin{aligned} & x^2+y^2-6 x-8 y+\left(25-a^2\right)=0 \\ & \text { Radius }=4=\sqrt{9+16+\left(25-a^2\right)} \\ & \Rightarrow \mathrm{a}=\pm 4 \end{aligned} $

Asked in: JEE Main 2013 (23 Apr Online)

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