## How to solve circle equation? ax^2 - 12x + ay^2 +16y = 25

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### How to solve circle equation? ax^2 - 12x + ay^2 +16y = 25

A circle is given as:
ax^2 - 12x + ay^2 +16y = 25

How to find 'a' ?

Thanks.
mitnord

Posts: 13
Joined: Mon Feb 23, 2009 11:17 pm

mitnord wrote:A circle is given as: ax^2 - 12x + ay^2 +16y = 25

How to find 'a' ?

I'm not sure what they're expecting from you here. Without additional information, surely there are infinitely-many possible values for "a"...?

You have:

. . . . .$ax^2\, -\, 12x\, +\, ay^2\, +\, 16y\, =\, 25$

Completing the square gives:

. . . . .$a\left(x^2\, -\, \frac{12}{a}x\right)\, +\, a\left(y^2\, +\, \frac{16}{a}y\right)\, =\, 25$

. . . . .$a\left(x^2\, -\, \frac{12}{a}x\, +\, \frac{36}{a^2}\right)\, +\, a\left(y^2\, +\, \frac{16}{a}y\, +\, \frac{64}{a^2}\right)\, =\, 25\, +\, a\left(\frac{36}{a^2}\right)\, +\, a\left(\frac{64}{a^2}\right)$

. . . . .$a\left(x\, -\, \frac{6}{a}\right)^2\, +\, a\left(y\, +\, \frac{8}{a}\right)^2\, =\, 25\, +\, \frac{100}{a}$

. . . . .$\left(x\, -\, \frac{6}{a}\right)^2\, +\, \left(y\, +\, \frac{8}{a}\right)^2\, =\, \frac{25}{a}\, +\, \frac{100}{a^2}\, =\, \frac{25a\, +\, 100}{a^2}$

But without some sort of restrictions on the circle, more information on "a", or something else, I see no way to find "the" value of "a".