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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.

Feb 21, 2018

#1
+7612
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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

Feb 21, 2018
#2
+7612
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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

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)

Feb 21, 2018
edited by hectictar  Feb 21, 2018