Minimum value . . .

Geometry Level 3

If real numbers $x$ and $y$ satisfy $(x+5)^2 + (y-12)^2=14^2$ , what is the minimum value of $x^2+y^2$ ?

\sqrt{3}

1

2

\sqrt{2}

2 solutions

Aryan Sanghi
Aug 17, 2020

This is equivalent to finding minimum distance of origin $O(0,0)$ from circle $(x+5)^2 + (y-12)^2 = 14^2$ as $\sqrt{x^2 + y^2} = \text{Distance of point }P(x, y) \text{ from origin}$ .

Now, center of circle $C = (-5, 12)$ and radius $R = 14$

$OC = \sqrt{(-5)^2 + 12^2} = 13$

Minimum distance $d = |OC - R|$

$\boxed{d = 1}$

Therefore, $\color{#3D99F6}{\boxed{min(x^2 + y^2) = 1}}$

I knew it was something related to circle, but didn't know what. Thanks(+3)!

Vinayak Srivastava - 9 months, 4 weeks ago

Thanku. :)

Aryan Sanghi - 9 months, 4 weeks ago

LOL I deleted the comment just now, I thought you didn't get notification, so I was writing again.

I understood it from my brother, thank you! I am making a problem, I will post it in the discussion.

Vinayak Srivastava - 9 months, 4 weeks ago

$x^2+y^2$ is not the distance but the square of it. Accidentally the minimum of the distance and square of it match, because both are equal to $1$ . You have to rectify in that. Never mind. Keep it up. You used coordinate geometry, while I have used trigonometry.

A Former Brilliant Member - 9 months, 4 weeks ago

I have rectified that sir. Thanku for appreciation sir. :)

Aryan Sanghi - 9 months, 4 weeks ago

Now find the maximum of $x+y$ .

A Former Brilliant Member - 9 months, 4 weeks ago

@A Former Brilliant Member – $max(x^2 + y^2) = (OC + R)^2 = 729$ . :)

Aryan Sanghi - 9 months, 4 weeks ago

@Aryan Sanghi – Not $x^2+y^2$ but $x+y$

A Former Brilliant Member - 9 months, 4 weeks ago

@A Former Brilliant Member – Ohk, that requires trigonometry sir. You're right. :)

Aryan Sanghi - 9 months, 4 weeks ago

A Former Brilliant Member
Aug 16, 2020

The substitutions $x=14\cos α-5,y=14\sin α+12$ satisfy the given equation.

So, $x^2+y^2=365+364\sin \left (α-\tan^{-1} \frac{5}{12}\right )$

Hence, $1\leq x^2+y^2\leq 729$ (since $-1\leq \sin (α-\tan^{-1} \frac {5}{12})\leq 1$ )

Therefore, the minimum value of $x^2+y^2$ is $\boxed 1$ when $x\approx 0.3846,y\approx -0.923$ .

Sir, see to my solution. :)

Aryan Sanghi - 9 months, 4 weeks ago

Minimum value . . .

2 solutions

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