anyone good at math???- honors algebra 2?

Question

ok. i have to find the 2 points that the line and the circle intersect.
x^2+y^2=36
and
-x+y=1

~does anyone know how to solve this algebraically? thank ya...

Answer 1

Solve this pair of equations simultaneously:

(equation 1) x^2+y^2=36
(equation 2) -x+y=1

So from equation 2, we see that y = x+1
substitute this 'y' into equation 1 (that is, everytime you see 'y' in equation 1, replace it with 'x+1')
Equation 1 becomes:
x^2 + (x+1)^2 = 36
x^2 + (x+1)(x+1) = 36
x^2 + (x^2 + 2x + 1) = 36
2x^2 + 2x + 1 = 36
subtract 36 from both sides (also called bringing the 36 onto the left side...need to change signs)

2x^2 + 2x - 35 = 0
Use the quadratic formula, with
a=2
b=2
c=-35
and you'll get:
x = -4.7125 or x = 3.7125
sub these values into equation 1 or 2 to find their corresponding y values:
y = 1 + x
so y = -3.7125 or y = 4.7125

So the points of intersection are:

(-4.7125, -3.7125) and (3.7125, 4.7125)

Answer 2

The easiest way is to use substitution.

-x + y = 1
y = x + 1

x² + y² = 36

Substitute for y in the equation for the circle.

x² + (x + 1)² = 36
x² + x² + 2x + 1 = 36
2x² + 2x - 35 = 0
x = [-2 ± √(2² + 4*2*35)] / (2*2) = (-2 ± √284)/4
x = (-2 ± 2√71)/4 = (-1 ± √71)/2

Now plug into the equation of the line.

y = x + 1

y = (-1 - √71)/2 + 1 = (1 - √71)/2

y = (-1 + √71)/2 + 1 = (1 + √71)/2

So the two points of intersection are:

(x,y) = [(-1 - √71)/2, (1 - √71)/2]
and
(x,y) = [(-1 + √71)/2, (1 + √71)/2]

Answer 3

You should start by solving both expressions for either x or y:

y^2=36-x^2
y=(36-x^2)^1/2 (same as squareroot(36-x^2))

and the second equation:

y=x+1

Now I guess we have to make one equation equal the other to find out where the values of x equal eachother:

x+1=(36-x^2)^1/2

Now we can square both sides to get rid of the root:

(x+1)^2=36-x^2 [(x+1)^2=x^2+2x+1]
x^2+2x+1=36-x^2 (bring everything to one side)
x^2+x^2+2x+1-36=0
2x^2+2x-35=0 (now use quadratic formula to solve)

x=[-b+-(b^2-4ac)^1/2]/2a (quadratic formula which I'm sure you know)

x=[-2+-(4-4(2)(-35))^1/2]/4
x=[-2+-(4+280)^1/2]/4

So:
x=-1/2+[(284)^1/2]/4
x=-1/2-[(284)^1/2]/4

At these two x values the line and the circle intercept.

I hope I helped! :)

Answer 4

Sure!
redefine the line
y = 1 + x

substitute into the first equation:

x ^ 2 + ( 1 + x ) ^ 2 = 36

x^2 + x^2 + 2x + 1 = 36

regroup

2 x^2 + 2x - 35 = 0

use quadratic formula, hopefully you'll get 2 solutions!!!

Answer 5

So the first is a circle that starts at 0,0 with radius of 6. Solve -x+y=1 which is y=x+1

Now do a substitution.
So x^2+(x+1)^2 = 36, solve for x your answers are x=-4.713 and 3.713

Answer 6

It is an honor.
Substitute y=x+1 from the 2nd eq into the first, and you get x^2+(x+1)^2=36.
2x^2+2x-35=0

Answer 7

x^2 + y^2 = 36 <--equation 1
-x + y = 1 <--equation 2
equation 2 --> y = 1 + x
substitute into equation 2
equation 2 --> x^2 + (1 + x)^2 = 36
x^2 + x^2 + 2x + 1 - 36 = 0
2x^2 + 2x -35 = 0

Answer 8

ohk..this is really simple....just substitute the value of y or x in the circle s equation and u would get it.
x^2+y^2=36
x^2+(1+x)^2 =36
2x^2+1+2x=36
solve the quadratic equation,
2x^2+2x -35=0

Answer 9

substitute

Solve for y= in equation 2 ===> y = x+1

now plug into equation1
x^2+y^2=36
x^2 + (x+1)^2 = 36
x^2 + x^2 + 2x + 1 = 36
2x^2 + 2x - 35 = 0

quadratic equation:
x=[ -b +/-sqrt{b^2-4ac} ] / 2a

a=2 b=2 c = -35

x = [-2 +/- sqrt(284)] / 4
x = [-2 +/- 2sqrt71] /4
x = [-1 +/-sqrt71] / 2

Plug x into line equation to get y
y = x+1
POINTS ARE:

( [-1 +sqrt71] / 2 , [1 +sqrt71] / 2)
and
( [-1-sqrt71] / 2 , [1-sqrt71] / 2)

Answer 10

x^2 + y^2 = 36 (1)
and
- x + y = 1 (2)
Solve (2) for y.
Substitute this expression for y in (1).
Expand the square, simplify and factor to get two values for x.
Substitute each value of x into (2) and solve for corresponding values of y.