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Find minimum value of $\sqrt{(a-p)^2 + (4-a-q)^2}$. Given $p^2 + q^2=1$.

My attempts

  1. I tried putting $p= \sin¢$ and $q = \cos ¢$ but in the end I got an equation with $\sin$ and $a$ which I couldn't solve further.

  2. I tried to think of it as distance formula with one pt. being $(p,q)$ lying on a circle and another point $(a, a-4)$ which is on a line $x-y=4$ so I plotted it and I got the answer as $2√2-1$. Is it correct?

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    $\begingroup$ Is $a$ a fixed value, or is it also variable? $\endgroup$
    – ConMan
    Commented Dec 8, 2023 at 5:49
  • $\begingroup$ a is a variable $\endgroup$
    – Harshita Jain
    Commented Dec 8, 2023 at 6:26
  • $\begingroup$ $(a-p)^2+(4-a-q)^2 = a^2-2ap+p^2+(4-a)^2-2(4-a)q+q^2= a^2-2ap+(4-a)^2-2(4-a)q+1= ...$, does this help? $\endgroup$
    – Dominique
    Commented Dec 8, 2023 at 8:14
  • $\begingroup$ @HarshitaJain: The point must not be $$(a,4-a)$$? and the line $x+y=4$? $\endgroup$
    – Khosrotash
    Commented Dec 8, 2023 at 8:25
  • $\begingroup$ +1 to your posting. Your analysis is accurate, valid and elegant. The closest point of form $~(a,4-a)~$ to the circle occurs at $~a=2,~$ and the closest point of intersection of this point to the circle is with point $~(x,y) = \displaystyle \left( ~\frac{1}{\sqrt{2}}, ~\frac{1}{\sqrt{2}} ~\right).~$ The distance is $~\displaystyle \sqrt{2 \times \left[ ~2 - \frac{1}{\sqrt{2}} ~\right]^2}.~$ This equals $~\displaystyle \sqrt{2} \times \left[ ~2 - \frac{1}{\sqrt{2}} ~\right] = \left[ ~2\sqrt{2} - 1 ~\right].$ $\endgroup$
    – user2661923
    Commented Dec 8, 2023 at 9:00

1 Answer 1

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The distance required is exactly the distance between $(p,q)$ and $(a,4-a)$. Since $(p,q)$ is on the unit circle with centre $(0,0)$, therefore the minimum value of $\sqrt{(a-p)^2 + (4-a-q)^2}$ is the shortest distance between the circle and line $y=4-x$ and equal to $$\frac{|0+0-4|}{\sqrt{1^2+1^2}}-1=2\sqrt 2-1 .$$

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