Circle problem

Given A circle with center $O$ and random point $P$ "on" the side of the circle . Let there be Point's $A$ and $B$ on the same diameter but not on the same point such that $OA = OB$ Prove (or disprove ) that $(PA)^2+(PB)^2$ is constant

#Geometry

Easy Math Editor

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Comments

Coordinates of points:

$Px = R \, cos \theta \\ Py = R \, sin \theta \\ Ax = -\alpha \\ Ay = 0 \\ Bx = \alpha \\ By = 0$

Squared distances:

$PA^2 = (R \, cos \theta + \alpha)^2 + (R \, sin \theta)^2 = R^2 + 2 \, \alpha \, R \, cos \theta + \alpha^2 \\ PB^2 = (R \, cos \theta - \alpha)^2 + (R \, sin \theta)^2 = R^2 - 2 \, \alpha \, R \, cos \theta + \alpha^2 \\ PA^2 + PB^2 = 2 R^2 + 2 \alpha^2 = \text{constant}$

Thus, there is no angular dependence. So for a particular $R$ and a particular $\alpha$ , $PA^2 + PB^2$ is constant

Steven Chase - 2 years, 3 months ago

But I think $\alpha$ is not constant and is changing.

Ram Mohith - 2 years, 3 months ago

I guess we just need to know what the word "constant" is supposed to mean here. I took it to mean "no angular variance". For varying $\alpha$ , the result will vary.

Steven Chase - 2 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$