Question

If $(a,0)$ is a point on a diameter segment 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

The correct option is A exactly one real root in $(-1,0]$

Given $x^2-4x-a^2=0$

$D=16-\mathrm{4.1.}(-a^2)$

$D=16+4a^2$

$\alpha +\beta =4$

$\alpha \cdot \beta =-a^2$

That mean one root is positive and other is negative

$-2\le a\le 2$

So, answer is $A$