# Question 59984

Feb 23, 2017

$\left(- \frac{18}{5} , \frac{1}{5}\right) , \left(3 , - 2\right)$

#### Explanation:

Solve using $\textcolor{b l u e}{\text{substitution}}$

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

$x + 3 y = - 3 \Rightarrow x = \textcolor{b l u e}{- 3 - 3 y} \to \left(1\right)$

$\Rightarrow {\left(\textcolor{b l u e}{- 3 - 3 y}\right)}^{2} + {y}^{2} = 13$

distributing the bracket using the FOIL method.

$\Rightarrow \left(9 + 18 y + 9 {y}^{2}\right) + {y}^{2} = 13$

simplify and equate to zero.

$\Rightarrow 10 {y}^{2} + 18 y - 4 = 0$

factorising the left side.

$2 \left(5 y - 1\right) \left(y + 2\right) = 0$

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]}