# Solve the equation ((2-1x)^2+(1-1y)^2)^(1/2)=10

## \left(\left(2-x\right)^2+\left(1-y\right)^2\right)^{\frac{1}{2}}=10

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$y^2-2y+x^2-4x=95$

## Step by step solution

Problem

$\left(\left(2-x\right)^2+\left(1-y\right)^2\right)^{\frac{1}{2}}=10$
1

Expanding the polynomial

$\sqrt{\left(2-x\right)^2+y^2-2y+1}=10$
2

Expanding the polynomial

$\sqrt{y^2-2y+1+x^2-4x+4}=10$
3

Add the values $4$ and $1$

$\sqrt{y^2-2y+x^2-4x+5}=10$
4

Removing the variable's exponent

$\left(y^2-2y+x^2-4x+5\right)^{div\left(1,\frac{1}{2}\right)\cdot \frac{1}{2}}=10^{div\left(1,\frac{1}{2}\right)}$
5

Divide $1$ by $\frac{1}{2}$

$\left(y^2-2y+x^2-4x+5\right)^{2\cdot \frac{1}{2}}=10^{2}$
6

Multiply $\frac{1}{2}$ times $2$

$\left(y^2-2y+x^2-4x+5\right)^{1}=10^{2}$
7

Calculate the power

$\left(y^2-2y+x^2-4x+5\right)^{1}=100$
8

Any expression to the power of $1$ is equal to that same expression

$y^2-2y+x^2-4x+5=100$
9

Moving the term $5$ to the other side of the equation with opposite sign

$y^2-2y+x^2-4x=100-5$
10

Subtract the values $100$ and $-5$

$y^2-2y+x^2-4x=95$

$y^2-2y+x^2-4x=95$

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