# How do you solve (x-3)^2+(y-5)^2=25 and x^2+(y-1)^2=100?

Aug 8, 2015

#### Answer:

Expand the squares for the 2 equations for circles, subtract to get a linear equation, substitute back into one of the circle equations:
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x , y\right) = \left(6 , 9\right)$

#### Explanation:

[1]$\textcolor{w h i t e}{\text{XXXX}}$${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25$
[2]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + {\left(y - 1\right)}^{2} = 100$

Expanding the square for [1]
[3]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 6 x + 9 + {y}^{2} - 10 y + 25 = 25$
Simplifying
[4]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 6 x + {y}^{2} - 10 y = - 9$

Expanding the square for [2]
[5]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + {y}^{2} - 2 y + 1 = 100$
Simplifying
[6]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + {y}^{2} - 2 y = 99$

Subtract [4] from [6]
[7]$\textcolor{w h i t e}{\text{XXXX}}$$6 x + 8 y = 108$
Simplify
[8]$\textcolor{w h i t e}{\text{XXXX}}$$y = \frac{54 - 3 x}{4}$

Substituting $\left(\frac{54 - 3 x}{4}\right)$ for $y$ in [2]
[9]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + {\left(\frac{54 - 3 x}{4} - 1\right)}^{2} = 100$
or
[10]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + {\left(\frac{50}{4} - \frac{3 x}{4}\right)}^{2} = 100$

Expanding the square
[11]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + \left(\frac{2500}{16} - \frac{300 x}{16} + \frac{9 {x}^{2}}{16}\right) = 100$

Multiply by 16
[12]$\textcolor{w h i t e}{\text{XXXX}}$$16 {x}^{2} + 2500 - 300 x + 9 {x}^{2} = 1600$

Simplifying into standard quadratic form
[13]$\textcolor{w h i t e}{\text{XXXX}}$$25 {x}^{2} - 300 x + 900 = 0$
Dividing by $25$
[14]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 12 x + 36 = 0$

Factoring
[15]$\textcolor{w h i t e}{\text{XXXX}}$${\left(x - 6\right)}^{2} = 0$$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow x = 6$

When $x = 6$ (substituting back into [8]

[16]$\textcolor{w h i t e}{\text{XXXX}}$$y = \frac{54 - 3 \left(6\right)}{4} = 9$