# System of Equations: Circle, Line

How to solve the system of circle and linear equations: 1 example and its solution.

## Examplex^{2} + y^{2} = 25, y = x + 1

Change the linear equation to [y = ...].

[y = x + 1] is already in that form.

Then put y = x + 1

into the circle x^{2} + y^{2} = 25.

Then x^{2} + (x + 1)^{2} = 25.

Substitution Method

(x + 1)^{2}

= x^{2} + 2⋅x⋅1 + 1^{2}

= x^{2} + 2x + 1

Square of a Sum: (a + b)^{2}

x^{2} + x^{2} = 2x^{2}

Move 25 to the left side.

Then 2x^{2} + 2x + 1 - 25 = 0.

+1 - 25 = -24

Divide both sides by 2.

Then x^{2} + x - 12 = 0.

Factor the right side

x^{2} + x - 12.

Find a pair of numbers

whose product is the constant term -12

and whose sum is the coefficient of the middle term +1.

-3⋅4 = -12

-3 + 4 = +1

Then (x - 3)(x + 4) = 0.

Factor a Quadratic Trinomial

To find the y value for this case,

put x = 3

into the linear equation y = x + 1.

Then y = 3 + 1 = 4.

x = 3

y = 4

So (3, 4) is the answer for case 1.

Case 2) x + 4 = 0

Then x = -4.

To find the y value for this case,

put x = -4

into the linear equation y = x + 1.

Then y = -4 + 1 = -3.

x = -4

y = -3

So (-4, -3) is the answer for case 2.

Case 1) (3, 4)

Case 2) (-4, -3)

Write these two points.

So (3, 4), (-4, -3) is the answer.

These are the graphs of

the circle [x^{2} + y^{2} = 25]

and the line [y = x + 1].

By solving the system,

you found the intersecting points:

(3, 4) and (-4, -3).