System of Equations: Circle, Line

Change the linear equation to [y = ...].
[y = x + 1] is already in that form.

Then put y = x + 1
into the circle x² + y² = 25.

Then x² + (x + 1)² = 25.

Substitution Method

(x + 1)²
= x² + 2⋅x⋅1 + 1²
= x² + 2x + 1

Square of a Sum: (a + b)²

x² + x² = 2x²

Move 25 to the left side.

Then 2x² + 2x + 1 - 25 = 0.

+1 - 25 = -24

Divide both sides by 2.

Then x² + x - 12 = 0.

Factor the right side
x² + 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

Solve the quadratic equation.

Case 1) x - 3 = 0
Then x = 3.

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 [x² + y² = 25]
and the line [y = x + 1].

By solving the system,
you found the intersecting points:
(3, 4) and (-4, -3).