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Let $a\in\mathbb{R}$. The number of distinct solutions $(x,y)$ that satisfy the system of equations $(x-a)^{2}+y^{2}=1$ and $x^{2}=y^{2}$ can only be _____.

Kenta S
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Keshav Balwant Deoskar
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  • First ask yourself how many way can $x^2$ equal $y^2$, find $x$ as a function of $y$ (or vice-versa) and substitute inside $(x-a)^2+y^2=1$, you'll find a quadratic equation and the number of its solutions will be a function of $a$ (depending on the sign of the discriminant), the number of $(x,y)$ depend on the number of solutions of this quadratic equation – Alessandro Mar 20 '20 at 14:29

1 Answers1

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I will say that

$$(x-a)^2 + y^2 = 1$$

is the equation of a circle of radius $r=1$ centered at $x=a$. The additional constraint that $x^2 = y^2$ implies that $y= \pm x$ which are two lines of $y-$intercept $0$ and slope $\pm1$. Now depending on the value of $a$, the circle can be quite far from the origin where it will not intersect the lines (a.k.a. no solutions). It can also be close to the origin where it can have either $2$ solutions or $4$ solutions. Again, the answer depends heavily on the value of $a$.

JacobCheverie
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