Question

If $(a,0)$ is an endpoint of a diameter of the circle $x^2+y^2=4$, then $x^2-4x-a^2=0$ has

• A. exactly one real root in $(-1,0]$
• B. exactly one real root in $[2,5]$
• C. distinct roots greater than $-1$
• D. distinct roots less than $5$

Answer 1

Answer: (A), (B), (C), (D)

distinct roots less than $5$

Since, $(a,0)$ is an endpoint of a diameter of the circle $x^2+y^2=4$.
$\therefore a^2+0^2=4$
$\Rightarrow a^2=4$

Given, $x^2-4x-a^2=0$
$\Rightarrow x^2-4x-4=0$
$\Rightarrow x=\frac{4\pm \sqrt{32}}{2}$
$\Rightarrow x=2\pm 2\sqrt{2}$
$\Rightarrow x=2(1\pm \sqrt{2})$