Question

Prove: If a chord of circle $x^2+y^2=8$ makes equal intercepts of length 'a' on the coordinate axes then $|a|<4$

Answer 1

The chord of the circle let $x{x}_1+y{y}_1-8=0$

Writing in the intercept form we would get,

$\Rightarrow \frac{x}{\frac{y_1}{x_1}}+y(-1)=\frac{-8}{y_1}$

and given intercept are equal to $a$

Hence, we get

$\Rightarrow x_1=2$ and $y_1=2$

$\Rightarrow a=$ $\sqrt{(4+4)}$

$2\sqrt{2}$ which is less than $4.$

Hence, solved.