# To find minimum value of an equation

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?

• Is $a$ a fixed value, or is it also variable? Commented Dec 8, 2023 at 5:49
• a is a variable Commented Dec 8, 2023 at 6:26
• $(a-p)^2+(4-a-q)^2 = a^2-2ap+p^2+(4-a)^2-2(4-a)q+q^2= a^2-2ap+(4-a)^2-2(4-a)q+1= ...$, does this help? Commented Dec 8, 2023 at 8:14
• @HarshitaJain: The point must not be $$(a,4-a)$$? and the line $x+y=4$? Commented Dec 8, 2023 at 8:25
• +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 $~\displaystyle \sqrt{2} \times \left[ ~2 - \frac{1}{\sqrt{2}} ~\right] = \left[ ~2\sqrt{2} - 1 ~\right].$ Commented Dec 8, 2023 at 9:00

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 $$\frac{|0+0-4|}{\sqrt{1^2+1^2}}-1=2\sqrt 2-1 .$$