Try this beautiful Problem on Algebra based on Problem on Curve from AMC 10 A, 2018. You may use sequential hints to solve the problem.

## Curve- AMC 10A, 2018- Problem 21

Which of the following describes the set of values of $a$ for which the curves $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ in the real $x y$ -plane intersect at

exactly 3 points?

- $a=\frac{1}{4}$
- $\frac{1}{4}<a<\frac{1}{2}$
- $a>\frac{1}{4}$
- $a=\frac{1}{2}$
- $a>\frac{1}{2}$

**Key Concepts**

Algebra

greatest integer

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-14

#### Check the answer here, but try the problem first

$a>\frac{1}{2}$

## Try with Hints

#### First Hint

We have to find out the value of \(a\)

Given that $y=x^{2}-a$ . now if we Substitute this value in $x^{2}+y^{2}=a^{2}$ we will get a quadratic equation of $x$ and \(a\). if you solve this equation you will get the value of \(a\)

Now can you finish the problem?

#### Second Hint

After substituting we will get $x^{2}+\left(x^{2}-a\right)^{2}$=$a^{2} \Longrightarrow x^{2}+x^{4}-2 a x^{2}=0 \Longrightarrow x^{2}\left(x^{2}-(2 a-1)\right)=0$

therefore we can say that either \(x^2=0\Rightarrow x=0\) or \(x^2-(2a-1)=0\)

\(\Rightarrow x=\pm \sqrt {2a-1}\). Therefore

Now Can you finish the Problem?

#### Third Hint

Therefore \(\sqrt {2a-1} > 0\)

\(\Rightarrow a>\frac{1}{2}\)

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=y-8Ru_qKDxk