If the circle x^2+y^2-6x-8y+(25-a^2)=0 touches the axis of x, then a equals:

Question

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

Accepted Answer

Centre $(3,4)$, radius $r=\sqrt{9+16-(25-a^2)}=\sqrt{a^2}=|a|$. Tangency to the $x$-axis requires $r=|$y-coordinate of centre$|=4$:
$$|a|=4\Rightarrow a=\pm4.$$

Solutions · 1