Question

Question #59984

Answer 1

`(-18/5,1/5),(3,-2)`

Explanation:

Solve using `color(blue)"substitution"`

Rearrange the linear equation to give x in terms of y, then substitute into the equation of the circle.

`x+3y=-3rArrx=color(blue)(-3-3y)to(1)`

`rArr(color(blue)(-3-3y))^2+y^2=13`

distributing the bracket using the FOIL method.

`rArr(9+18y+9y^2)+y^2=13`

simplify and equate to zero.

`rArr10y^2+18y-4=0`

factorising the left side.

`2(5y-1)(y+2)=0`

solving for y gives.

`•5y-1=0rArry=1/5`

`•y+2=0rArry=-2`

Substitute these values for y into (1) to obtain corresponding
x-coordinates.

`• y=1/5to x=-3-3/5=-18/5to(-18/5,1/5)`

`• y=-2 to x=-3+6=3 to(3,-2)`

The graph illustrates, geometrically, the intersections of the linear equation with the circle.
graph{(x^2+y^2-13)(y+1/3x+1)=0 [-10, 10, -5, 5]}