# Question #59984

##### 1 Answer

Feb 23, 2017

#### Answer:

#### 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]}