Analytic Geometry – Problem 7: Determine the positional relationship between line and circle , and justify your conclusion

Given the circle $C:x^2+y^2=16$ and the line $l:y=mx+3+2m\,(m\in\mathbb R)$.

(1) Determine the positional relationship between line $l$ and circle $C$, and justify your conclusion.

(2) If $P(a,b)$ is any point on circle $C$, find the range of $a+2b$.

(1) Rewrite the line as $$y=m(x+2)+3.$$ Set $x+2=0\Rightarrow x=-2$, then $y=3$, so every such line passes through the fixed point $$M(-2,3).$$ Since $$(-2)^2+3^2=13<16,$$ point $M$ lies inside the circle. Any line through an interior point intersects the circle, so $l$ intersects circle $C$.

(2) Let $$t=a+2b.$$ Together with the circle equation, $$\begin{cases} a^2+b^2=16,\\ a+2b=t. \end{cases}$$ Eliminate $a$: $$(a+2b-2b)^2+b^2=16\Rightarrow 5b^2-4tb+t^2-16=0\quad(t=a+2b).$$ For real $b$, the discriminant must satisfy $$\Delta=16t^2-20(t^2-16)=320-4t^2\ge0.$$ Therefore $$-4\sqrt5\le t\le4\sqrt5.$$ So $$-4\sqrt5\le a+2b\le4\sqrt5.$$

(1) Line $l$ always intersects circle $C$. The key point is that every such line passes through $M(-2,3)$, and $M$ is inside $x^2+y^2=16$.

(2) Letting $t=a+2b$ and imposing the real-solution condition for the derived quadratic gives $$-4\sqrt5\le t\le4\sqrt5.$$ Hence the required range is $$-4\sqrt5\le a+2b\le4\sqrt5.$$