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# Problem on Curve | AMC 10A, 2018 | Problem 21 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?

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

Algebra

greatest integer

## Suggested Book | Source | Answer

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 $\sqrt {2a-1} > 0$

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

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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?

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

Algebra

greatest integer

## Suggested Book | Source | Answer

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 $\sqrt {2a-1} > 0$

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

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