Question

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

Answer 1

Expand the squares for the 2 equations for circles, subtract to get a linear equation, substitute back into one of the circle equations:
`color(white)("XXXX")` `(x,y) = (6,9)`

Explanation:

[1]`color(white)("XXXX")` `(x-3)^2+(y-5)^2=25`
[2]`color(white)("XXXX")` `x^2+(y-1)^2=100`

Expanding the square for [1]
[3]`color(white)("XXXX")` `x^2-6x+9 + y^2-10y+25 = 25`
Simplifying
[4]`color(white)("XXXX")` `x^2-6x+y^2-10y = -9`

Expanding the square for [2]
[5]`color(white)("XXXX")` `x^2 +y^2-2y+1 = 100`
Simplifying
[6]`color(white)("XXXX")` `x^2+y^2-2y= 99`

Subtract [4] from [6]
[7]`color(white)("XXXX")` `6x +8y = 108`
Simplify
[8]`color(white)("XXXX")` `y = (54-3x)/4`

Substituting `((54-3x)/4)` for `y` in [2]
[9]`color(white)("XXXX")` `x^2+((54-3x)/4 -1)^2 = 100`
or
[10]`color(white)("XXXX")` `x^2+(50/4-(3x)/4)^2 = 100`

Expanding the square
[11]`color(white)("XXXX")` `x^2 + (2500/16-(300x)/16+(9x^2)/16) = 100`

Multiply by 16
[12]`color(white)("XXXX")` `16x^2 +2500-300x+9x^2 = 1600`

Simplifying into standard quadratic form
[13]`color(white)("XXXX")` `25x^2-300x+900 = 0`
Dividing by `25`
[14]`color(white)("XXXX")` `x^2-12x+36 = 0`

Factoring
[15]`color(white)("XXXX")` `(x-6)^2 = 0` `color(white)("XXXX")` `rarr x=6`

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

[16]`color(white)("XXXX")` `y = (54-3(6))/4 = 9`