Question

If a chord of the circle $x^2+y^2=8$ makes equal intercepts $a$ on the coordinate axes, then

• A. $|a|<2$
• B. $|a|\le 2\sqrt{2}$
• C. $|a|<4$
• D. $|a|<\sqrt{2}$

Answer 1

The correct option is C $|a|<4$

Circle: $x^2+y^2=8=(2\sqrt{2}{)}^2$
centre,$C=(0,0);$ radius$,r=2\sqrt{2}$
Equation of chord is $\frac{x}{a}+\frac{y}{a}=1\Rightarrow x+y=a$
$\therefore$ perpendicular distance from $C$ to the above line $<$ radius
$\Rightarrow \frac{|a|}{\sqrt{1^2+1^2}}<2\sqrt{2}\Rightarrow |a|<4$