Question

Find all solutions to the following system of equations x+y=4 ; (x+3)²+(y-2)²=10. Illustrate with a graph.

Answer 1

Explanation:

`x+y=4`

`(x+3)^(2)+(y-2)^(2)=10`

From first equation

`x=4-y`

Substituting in the second equation

`(4-y+3)^(2)+(y-2)^(2)=10\\\Rightarrow (7-y)^2+y^2+4-4y=10\\\Rightarrow 49+y^2-14y+y^2+4-4y=10\\\Rightarrow 2y^2-18y+43=0`

Solving the equation we get

`y=(-\left(-18\right)+√(\left(-18\right)^2-4\cdot \:2\cdot \:43))/(2\cdot \:2), (-\left(-18\right)-√(\left(-18\right)^2-4\cdot \:2\cdot \:43))/(2\cdot \:2)\\\Rightarrow y=(9)/(2)+i(√(5))/(2), (9)/(2)-i(√(5))/(2)`

Hence the circle and the line will not intersect and have no solutions