Question

Find the points where the line y = x - 1 intersects the circle x2 + y2 = 13

Answer 1

Answer:

(-2,-3) and (3,2)

Step-by-step explanation:

sub in x-1 into y

x^2 + (x-1)^2 = 13

x^2 + (x-1)(x-1)=13

x^2 + x^2 -2x +1 = 13

2x^2 -2x-12=0

solve for x by factoring (quadratic formula, product sum etc..)

x= -2 and 3

plug in those values into y=x-1 and solve for y

Answer 2

Answer:

$\large \boxed{(-2,-3) \text{ and } (3, 2)}$

Step-by-step explanation:

1. Solve the equations for x

$\begin{array}{lrcll}(1) & y & = & x - 1 & \\(2) & x^{2} + y^{2} &= &13& \\& x^{2} + (x - 1)^{2} &= & 13& \text{Substituted (1) into (2)}\\& x^{2} + x^{2} -2x +1 & = & 13 & \text{Squared (x - 1)} \\&2x^{2} -2x +1 & = & 13 & \text{Combined like terms} \\\end{array}$

$\begin{array}{lrcll}&2x^{2} -2x - 12 & = & 0 & \text{Subtracted 13 from each side}\\&x^{2} - x - 6 & = & 0 & \text{Divided each side by 2}\\& (x - 3)(x + 2) & = & 0 & \text{Factored the left-hand side} \\& x - 3 = 0 & \text{or} & x + 2 = 0 & \text{Applied zero product rule} \\& \mathbf{x = 3} & \text{or} & \mathbf{x = -2} & \text{Solved each equation separately} \\\end{array}$

2. Calculate the corresponding values of y

Insert the values into equation (1)

(a) x = 3

y = 3 - 1 = 2

One point of intersection is (3, 2).

(b) x = -2

y = -2 - 1 = -3

The second point of intersection is (-2, -3).

$\text{The line intersects the circle at \large \boxed{\mathbf{(-2,-3)} \text{ and } \mathbf{(3, 2)}} }$

The diagram shows the intersection of the two graphs.