Circle - maximum and minimum values

Question

If $x$ and $y$ are satisfying $x^2+y^2-6x-8y=0$ and $a=max({x+y})$ and $b=min({x-y})$ then what is the value of $a+b$?

My try: I drew the graph of the equation and found that the circle passes from the point$(0,0)$.Then on finding the center as $(3,4)$, I found the coordinates of the diametrically opposite end as $(6,8)$[which according to me should be $a=max({x+y})$], and it came out to be 14, and since the circle is passing through the point$(0,0)$, $b=min({x-y})$ should be $0$.But that's not correct.

$$ $$ Also, is there any way this can be done using parameters like $x = rcos (\theta)$ and $y = rsin (\theta)$?

Answer 1

Note that the equation of the circle is $(x-3)^2+(y-4)^2=5^2$, which can be parametrised as $x=3+5\cos\theta, y=4+5\sin\theta$. Now $x+y=7+5(\cos\theta+\sin\theta)=7+5\sqrt{2}\sin(\theta+\frac{\pi}{4})$ and so $a=7+5\sqrt{2}$ and similarly $x-y=-1+5(\cos\theta-\sin\theta)=-1+5\sqrt{2}\sin(\frac{\pi}{4}-\theta)$ and so $b=-1-5\sqrt{2}$

Answer 2

$x^2+y^2-6x-8y=0\Rightarrow (x-3)^2+(y-4)^2 = (5)^2$

Now let $x+y=k\;,$ Then $(x-3)+(y-4) = k-7$

So Using $$2(A^2+B^2)\geq (A+B)^2\cdots \cdots (1)$$

Setting $A=x-3$ and $B=y-4$ in eqn $(1)$

So $$2\bigg((x-3)^2+(y-4)^2\bigg)\geq (x-3+y-4)^2$$

So $$(k-7)^2\leq (2\sqrt{5})^2\Rightarrow k\leq \underbrace{7+2\sqrt{5}}_{\bf{a}}$$

Setting $x-y=b\;,$ Then $(x-3)-(y-4) = l+1$

Setting $A=x-3$ and $B=y-4$ in eqn $(1)$

So $$2\bigg((x-3)^2+(y-4)^2\bigg)\geq \bigg((x-3)-(y-4)\bigg)^2$$

So $$(l+1)^2\geq (2\sqrt{5})^2\Rightarrow l\leq \underbrace{-1-2\sqrt{5}}_{\bf{b}}$$

So $$a+b =7+2\sqrt{5}-1-2\sqrt{5} = 6$$

Answer 3

Let $x+y=2u,x-y=2v$

$$\implies(2u-7)^2+(2v+1)^2=50$$

$$\implies2u-7\le\sqrt{50}\implies2u\le7+\sqrt{50}$$

and $$2v+1\le\sqrt{50}\iff-\sqrt{50}-1\le2v\le\sqrt{50}-1$$

$$2u_{\text{max}}+2v_{\text{min}}=7+\sqrt{50}+(-\sqrt{50}-1)$$