JEE Main 2013 (Online) 23rd AprilQuestion 1
If the circle $x^{2}+y^{2}-6x-8y+\left(25-a^{2}\right)=0$ touches the axis of $x$, then $a$ equals :
Circle
Solution
Concept. A circle touches the $x$-axis when its radius equals $|k|$. Formulas used. (1) $r=\sqrt{g^2+f^2-c}$.(2) Touches $x$-axis $\iff r=|k|$.Step 1 - Use (1). Centre $(3,4)$, $r^2=9+16-(25-a^2)=a^2$. Step 2 - Use (2). $r=4\Rightarrow a^2=16\Rightarrow a=\pm4$. Answer: B