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I recently gave a test and had this question there . To be honest I was confused on seeing this question. I was unable to even deduce the approach. I am a student studying in class 11 so I don't have access to desmos during tests.

insipidintegrator
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Karan Suthar
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2 Answers2

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Rewrite the equation as $$(x^2+y^2)^2-16x^2+6x^2+6y^2+9=0$$ which, using the difference of two squares, is equivalent to $$(x^2-4x+y^2)(x^2+4x+y^2)+6(x^2+y^2)+9=0.$$ Then rewrite $6(x^2+y^2)$ as $3(x^2-4x+y^2)+3(x^2+4x+y^2)$ to enable us to factorise.

Now let $a=x^2-4x+y^2,b=x^2+4x+y^2$. Then we have $ab+3a+3b+9=0$, which factorises as $(a+3)(b+3)=0$. Hence the equations of the circles are $a+3=0$ and $b+3=0$, or $$(x-2)^2+y^2=1, (x+2)^2+y^2=1.$$

A. Goodier
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    That was a really good explanation for a really good question. Thanks a lot – Karan Suthar Dec 19 '22 at 12:00
  • I just wanted to just trouble you with one thing how do you build the approach to solve this kind of problem – Karan Suthar Dec 19 '22 at 12:00
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    Knowing tricks for factorising partly comes with experience but the fact it's a pair of circles here makes it easier, because we know both brackets will have $x^2$ and $y^2$, so most likely it will help to factorise the first three terms. But then, to be able to combine the rest of the terms with the $(x^2+y^2)^2$, it will help to have the same coefficient of $x^2$ and $y^2$ which is why I split up the $-10x^2$. The more expressions you try to factorise, the easier it gets! – A. Goodier Dec 19 '22 at 12:17
  • Thanks for the tip. – Karan Suthar Dec 19 '22 at 12:27
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    The factorization process can be simplified by noting that $$\left(x^2+y^2 \right)^2+6\left(x^2+y^2 \right)+9-16x^2$$$$=\left(x^2+y^2 +3\right)^2-(4x)^2$$$$=\left(x^2+y^2-2x+3 \right)\left(x^2+y^2+2x+3 \right)$$ – Li Kwok Keung Dec 19 '22 at 12:57
  • @LiKwokKeung Good spot! – A. Goodier Dec 19 '22 at 13:55
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Let $P(x,y):=x^2 - 4 \cdot x + y^2 + 3$ and let $Q(x,y):=x^2 + 4 \cdot x + y^2 + 3$. Then, your question could be simply rewritten as:
"Are you able to show that $P(x,y) \cdot Q(x,y) = x^4 + y^4 + 2 \cdot x^2 \cdot y^2 − 10 \cdot x^2 + 6 \cdot y^2 + 9$?".

And the above (IMHO) would be a good excercise to perform by your own.

Marco Ripà
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