Find max. and min. value of 'r'

Question

Find $\max$ and $\min$ values of $r$ if
$$r^2 - 4x + 3 = 0$$
where $r= \sqrt{x^2+y^2}$

Tried differentiating but didn't get any result for max. value of $ \sqrt{x^2+y^2}$

score 5 · Accepted Answer · answered Dec 19 '13 at 16:18

5

We have $$x^2+y^2-4x+3=0\implies (x-2)^2+y^2=1$$

We can write $x=2+\cos\phi, y=\sin\phi$

$$\implies r^2=(2+\cos\phi)^2+(\sin\phi)^2=5+4\cos\phi$$

Now we know that for real $\phi$ , $$-1\le\cos\phi\le1$$

Can you take it from here?

answered Dec 19 '13 at 16:18

lab bhattacharjee

274,582

In a way, the first line is enough, the rest is obvious from the geometry. – André Nicolas Dec 19 '13 at 16:27
@AndréNicolas, I was going to add that Geometrically, this is the distance between the origin and any point on the circle – lab bhattacharjee Dec 19 '13 at 16:28

score 0 · Answer 2 · answered Dec 19 '13 at 16:21

If $r^2-4x+3=0$ then $y^2=4x-3-x^2=-(x-3)(x-1)$.

Since $y^2\ge0$, you must have $x \in [1,3]$.

Then, $r=\sqrt{4x-3}$, so $r$ is minimum when $\sqrt{4x-3}$ is minimum, that is $x=1$ (notice $x \to \sqrt{4x-3}$ is an incresaing function) and $y=0$, and $r=1$.

And $r$ is maximum when $\sqrt{4x-3}$ is maximum, thus $x=3$, $r=\sqrt{4\cdot3-3}=3$ and $y=0$.

Find max. and min. value of 'r'

2 Answers2