Using linear equations

[Dev2011]

For what values of b will the line y=2x+b be tangent to the circle x[sup:1z6w1k0y]2[/sup:1z6w1k0y]+y[sup:1z6w1k0y]2[/sup:1z6w1k0y]=9?
A) Use the linear equation to eliminate y from the equation of the circle, leaving an equation involving x and b.
B) If the line is to be tangent to the circle, then the equation you wrote in part A must have exactly one solution for x. Use this fact to write an equation involving the discriminant of the equation in part A.
C) Solve the equation you wrote in part B for b. (it should not contain x)

In beginning to solve this problem, I plugged (2x+b) in for y in the equation of the circle, eventually leaving me with 3x[sup:1z6w1k0y]2[/sup:1z6w1k0y]+4xb+b[sup:1z6w1k0y]2[/sup:1z6w1k0y]=9.

Now I am trying to move on from there, and I'm stuck. I think I should solve for x, but I'm not sure how to do that. I also don't understand what the discriminant of the equation in part A would be.

 

[mmm4444bot]

… In beginning to solve this problem, I plugged (2x+b) in for y in the equation of the circle, eventually leaving me with 3x[sup:360sd41t]2[/sup:360sd41t]+4xb+b[sup:360sd41t]2[/sup:360sd41t]=9.

You're on the right track, but the leading coefficient is not 3.

Check your work.

Given y = Ax^2 + Bx + C, the discriminant is B^2 - 4AC.

In this exercise, the symbol b is a constant.

For the equation in part A to have one solution, the discriminant must equal zero.

That's how you find the value of b.

After you correct your work in part A, please show us what you come up with for the discriminant, if you would like continued help.

 

[Dev2011]

Thank you. You're right. The leading coefficient should be 5.

Regarding the discriminant, I really don't understand what to plug into the equation, B[sup:2kp8e6on]2[/sup:2kp8e6on]-4AC=0.
I got as far as 16-4(5)...but i don't know what should go in for C. Should it be 1, or the variable b, or am I just missing something obvious?

 

[mmm4444bot]

5x^2 + 4xb + b^2 = 9

We always need to put quadratic equations into standard form, before using the quadratic formula or the discriminant. Otherwise, how would we know the values of A, B, and C? So, set the equation equal to zero.

5x^2 + 4xb + b^2 - 9 = 0

I told you that b is a constant. So, everything in this equation represents constants, except for the symbol x, which is the variable. I'll insert some unnecessary parentheses, in order to emphasize the coefficients A, B, and C for you.

(5)x^2 + (4b)x + (b^2 - 9) = 0

Can you now write the equation: discriminant = zero, and solve for b?

 

[Dev2011]

That would make the discriminant 4b[sup:1989apjo]2[/sup:1989apjo]-4(5)(b[sup:1989apjo]2[/sup:1989apjo]-9)=0

If my math is right, that makes b equal to the square root of 45.

 

[mmm4444bot]

That would make the discriminant 4b[sup:3cp1jd5x]2[/sup:3cp1jd5x]-4(5)(b[sup:3cp1jd5x]2[/sup:3cp1jd5x]-9) …

If my math is right, that makes b equal to the square root of 45.

It looks, to me, like you did not type what you mean.

The discriminant is (4b)^2 - 4(5)(b^2 - 9).

I think that your math is partially correct. This is a quadratic polynomial. It has two roots.

sqrt(45) is only one of them.

Also, some instructors like to see radicals simplified. Perhaps, your instructor does not care, but I would not accept the expression sqrt(45).