How many solutions exist for this system of equations y=x-3 y=x^2-6x+10
2 answers:
You might be interested in
Answer:
r =
Step-by-step explanation:
So we need to complete the square for both parts of the equation
First though we can add the 6 to the other side so we have x² + 2x + y² + 4y = 6
So first we can complete the square for x² + 2x
To do so we need to use to figure out the number we need to add to both sides
In this case our b is 2, so substituting this in we get
Here we add 1 to both sides and now we have x² + 2x + 1 + y² + 4y = 6 + 1
Now we can follow the same steps to complete the square for y² + 4y
Here our b is 4, so substituting this in we get
Now we add 4 to both sides and now we have x² + 2x + 1 + y² + 4y + 4 = 6 + 1 + 4
Now condensing everything we have (x + 1)² + (y + 2)² = 11
The formula for a circle is (x - h)² + (y - k)² = r²
In our equation we have r² = 11
To find the radius we need to take the square root of both sides to get r =
