
Neal D. answered 08/13/16
Tutor
4.9
(1,417)
Precalculus Knowledgeable
( x + 2 )2 + ( y - 1 )2 = 32 ; y = x2 + 4x + 7
You have a circle and a parabola; you are looking for their intersection;
they can intersect in the following # of points: 0 , 1, 2, 3, or 4
You need to solve the two equations simultaneously
You could substitute for one of the variables in one of the equn.
Looking for an easy way, using:
y = x2 + 4x +7
y - 7 = x2 + 4x
Complete the square making sure you do not change the value
of the right side of equation
y - 7 = ( x + 2 )2 - 4
y - 3 = ( x +2 )2
Now you can substitute this value for ( x + 2 )2 into the equation for
the circle
y - 3 + ( y - 1 )2 = 32
y -3 + y2 -2y + 1 = 9
y2 -y -11 = 0
y2 - y = 11
( y - .5)2 = 11 + .25
y - .5 = ± √11.25
y = .5 ± √ 11.25
Substitute this value into either equation a solve for x; you should end
up with either 2 or 4 different points
When they ask for answers to be in exact form, they want them to stay
with the "√ " sign