Line not intersecting circle, maximum value of expression involving radius

Question

If line $y+x=2$ do not intersect any member of circles $x^2 + y^2 -ax = 0$ at two distinct points where a is parameter, then maximum value of $|a + 4|$.

My try: Since the line does not intersect the circle at 2 distinct points therefore the distance of line from center of circle is greater than or equal to the radius of the circle

Circle is $x^2 + y^2 -ax = 0$ therefore its radius is $a$ and center is $(a, 0)$

Distance of line $x^2 + y^2 -ax = 0$ from center of circle is given by $\frac{|a - 2|}{\sqrt2}$

Therefore we get the inequality $$\frac{|a - 2|}{\sqrt2} \ge a$$

But I can't find a way to solve it further to obtain the maximum value of |a + 4|

Answer 1

Use substitution to solve the system of equations: \begin{align*} \begin{cases} x^2 + y^2 - ax = 0 \\ y + x = 2 \end{cases} &\implies x^2 + (2 - x)^2 - ax = 0 \\ &\implies x^2 + (4 - 4x + x^2) - ax = 0 \\ &\implies 2x^2 + (-a - 4)x + 4 = 0 \\ \end{align*} But since there is at most one intersection point, we know that this equation must have at most one solution so that the discriminant of the LHS is either negative or zero: \begin{align*} (-a - 4)^2 - 4(2)(4) &\leq 0 \\ (-(a + 4))^2 - 32 &\leq 0 \\ (a + 4)^2 &\leq 32 \\ \sqrt{(a + 4)^2} &\leq \sqrt{32} \\ |a + 4| &\leq 4\sqrt{2} \end{align*}