The X is 1,0, and 1
The g(x) is 22,20, and 18
I’m trying to study and i just cant figure this one out. Thanks to anyone who is willing to help.
5 Answers

x² + y² = 100
center (0, 0)
radius = 10
From the (x, g(x)) data, the slope of g(x) is 2 and the yintercept is 20.
g(x) = 2x – 20
perpendicular from center of circle to g(x):
y = (½)x
point of intersection: (8, 4)
distance between center of circle and (8, 4) = √(8² + 4²) = √82 < 10, so point of intersection is inside the circle. circle and g(x) intersect, although not at (8, 4). But they didn’t ask where.

If the line passes through (1, 22), (0, 20), and (1, 18)
Then the yintercept is b = 20.
The slope is m = (18 – (20)) / (1 – 0) = 2
So g(x) = mx + b = 2x – 20
CHECK:
g(1) = 2(1) – 20 = 22
g(0) = 2(0) – 20 = 20
g(1) = 2(1) – 20 = 18
So we need to solve
y = 2x – 20 and x^2 + y^2 = 100
x^2 + (2x – 20)^2 = 100
x^2 + 4x^2 80x + 400 = 100
5x^2 – 80x + 300 = 0
5(x^2 – 16x + 60) = 0
(x – 6)(x – 10) = 0
x = 6 or x = 10
If x = 6, y = 2(6) – 20 = 8
If x = 10, y = 2(10) – 20 = 0
So the two curves intersect at the points
(6, 8) and (10, 0)

g(x): (1,22),(0,20),(1,18)
=> g(x) = y = 2x – 20
Plug into equation of circle:
(2x – 20)^2 + x^2 = 100
4x^2 – 80x + 400 + x^2 – 100 = 0
5x^2 – 80x + 300 = 0
x^2 – 16x + 60 = 0
(x – 10)(x – 6) = 0
x = 10 or x = 6
=> y = 0 or y = 8
So, they intersect in points (10,0) and (6,8)

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RE:
Circle graph, y^2+x^2=100, and a linear function g(x). Will they intersect?
The X is 1,0, and 1
The g(x) is 22,20, and 18
I’m trying to study and i just cant figure this one out. Thanks to anyone who is willing to help.

Your g(x) is 22, 20, 18 does not make sense.
There should be some form of linear eq’n, such as g(x) = mx + c
y^2 + x^2 = 10^2
This is a circle centred on (0,0) The origin, with a radius of ’10’.
On the coordinate axes the four intersections, with the circle and the coordinate axes, are ; – (10,0) , (0,10) , (10, 0) & (0, 10).