MATH SOLVE

2 months ago

Q:
# Christopher is analyzing a circle, y^2 + x^2 = 121, and a linear function g(x). Will they intersect? A. Yes, at the positive x coordinates. B. Yes, at negative x coordinates. C. Yes, at the negative and positive x coordinates. D. No, they will not intersect

Accepted Solution

A:

Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.

g(x)=-2x+12 Β (from given table, find slope and y-intercept)

We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.

To show that this is the case,Β

substitute g(x) into the circle

x^2+(-2x+12)^2=121

x^2+4x^2-2*2*12x+144-121=0

5x^2-48x+23=0

Solve using the quadratic formula,

x=(48 ± √ (48^2-4*5*23) )/10

=0.5058 or 9.0942

So both solutions are real and both have positive x-values.

g(x)=-2x+12 Β (from given table, find slope and y-intercept)

We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.

To show that this is the case,Β

substitute g(x) into the circle

x^2+(-2x+12)^2=121

x^2+4x^2-2*2*12x+144-121=0

5x^2-48x+23=0

Solve using the quadratic formula,

x=(48 ± √ (48^2-4*5*23) )/10

=0.5058 or 9.0942

So both solutions are real and both have positive x-values.