Substitute y=x+1 into the equation that defines the circle.
(3,4) and (- 4,- 3)
Practice makes perfect
We will use the Substitution Method to find the point(s) at which the circle and the line intersect.
x^2+y^2=25 & (I) y=x+1 & (II)
Since the second equation is already solved for y, we will substitute y=x+1 into the first equation.
Let's focus on the first equation for a moment. We want to factor it and then use the Zero-Product Property. First, we will factor out the Greatest Common Factor (GCF) of the polynomial on the left-hand side.
To factor the expression on the left-hand side, we have to find a pair of factors of - 12 that has a sum of 1.
Factors of - 12
Sum of Factors
- 1 and 12
11
- 2 and 6
4
- 3 and 4
1 âś“
Now we can rewrite the left-hand side of the equation in the factored form.
x^2+x-12=0 ⇔ (x-3)(x+4)=0
Finally, we will use the Zero-Product Property.
We have that x=3 or x=- 4. Next, we will use the second equation, y=x+1, to find the corresponding y-values.
x
y=x+1
y
3
y= 3+1
4
- 4
y= - 4+1
- 3
The solutions of the system are (3,4) and (- 4,- 3). These are the points of intersection of the circle and the line. Below we have included a graph representing our system.
