1.For what value of $c$ will the circle with equation $x^2 - 10x + y^2 + 6y + c = 0$ have a radius of length 1?
2.Two parabolas are the graphs of the equations $y=2x^2-7x+1$ and $y=8x^2+5x+1$. Give all points where they intersect. List the points in order of increasing $x$-coordinate, separated by semicolons.
+9464 1.
x2 - 10x + y2 + 6y + c = 0 Let's get this equation into standard form.
Subtract c from both sides of the equation.
x2 - 10x + y2 + 6y = 0 - c Add 25 and 9 to both sides to complete the squares on the left.
x2 - 10x + 25 + y2 + 6y + 9 = 0 - c + 25 + 9 Factor each perfect square trinomial on the left.
(x - 5)2 + (y + 3)2 = 0 - c + 25 + 9
(x - 5)2 + (y + 3)2 = -c + 34
Now that the equation is in this form, we can see that
(the radius)2 = -c + 34 If the radius is 1 , then
12 = -c + 34
1 = -c + 34
c = 33
+9464 2.
We want to find the solutions to this system of equations:
y = 2x2 - 7x + 1
y = 8x2 + 5x + 1
8x2 + 5x + 1 = 2x2 - 7x + 1
Subtract 2x2 from both sides.
6x2 + 5x + 1 = -7x + 1
Add 7x to both sides.
6x2 + 12x + 1 = 1
Subtract 1 from both sides.
6x2 + 12x = 0
Factor 6x out of both terms on the left side.
6x(x + 2) = 0
Set each factor equal to zero.
6x = 0 or x + 2 = 0
x = 0 or x = -2
Using one of the equations, find y when x = 0 .
When x = 0 , y = 2(0)2 - 7(0) + 1 = 1
Find y when x = -2 .
When x = -2 , y = 2(-2)2 - 7(-2) + 1 = 8 + 14 + 1 = 23
The points of intersection are (0, 1) and (-2, 23)
