Question

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

Answer 1

Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.
g(x)=-2x+12   (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, 
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.

Answer 2

Yes, they will intersect at the positive x coordinates.

What is the equation of the circle?

The standard equation of a circle is: (x-h)^ 2 + (y-k) ^2 =r ^2 Where (h, k) is the coordinates of the center, and r is the radius of the circle.

Christopher is analyzing a circle, y^2 + x^2 = 121, and a linear function g(x). Will they intersect?

The given equation of the circle is;

$\rm x^2+y^2=121\\\\x^2+y^2=11^2$

The circle with center at (0,0) and radius r = 11 units and g(x) points (-1,14)  (0,12)  (1,10).

The linear function is;

$\rm y=mx+b$

From the graph on drawing all the co-ordinates lies in positive x and y axis.

Hence, they will intersect at the positive x coordinates.

Learn more about circle here;

https://brainly.com/question/9821926

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