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

Since $(a, 0)$is a point on the diameter along x-axis

$\therefore -2\le a\le 2\Rightarrow 0\le a^2\le 4$

Let $f\left(x\right)=x^2-4x-a^2$ Clearly, discriminant of $f\left(x\right) = 0$ is positive. So, it has real and distinct roots

Now, $f\left(0\right)f(-1)=\left(-a^2\right)\left(5-a^2\right)<0\left[\because 0\le a^2\le 4\right]$

$\Rightarrow f\left(x\right)=0$ has a root in $(-1, 0]$

Also, $f\left(2\right)f\left(5\right)=\left(-4-a^2\right)\left(5-a^2\right)<0$

$\Rightarrow f\left(x\right)=0$ has a root in $[2, 5]$

Hence, all the options are correct