Pre-Calc: line thru (1,1), center of x^2-4x+y^2-8y+16=0

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caseygaspar

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Hey guys I was having trouble in my pre-calc homework.

Find the equation for the following lines satisfying the given conditions. Write the answers in the form of y=mx+b

Passes through (1,1) and the center of the circle x^2-4x+y^2-8y+16=0
I got it down to find that the center was (2,4) but then I got stuck there.
Please I really need help because I have a big test tomorrow!!
Thanks!
 
Re: Pre-Calc help

You have the worst part over. You have the correct circle center.

Now, you have two points the line passes through. (1,1) and (2,4).

Slope = \(\displaystyle \frac{4-1}{2-1}=3\)

Now, use any one of the data points, solve for b and you're done. I'd use (1,1)

\(\displaystyle y=3x+b\)
 
caseygaspar said:
Hey guys I was having trouble in my pre-calc homework.

Find the equation for the following lines satisfying the given conditions. Write the answers in the form of y=mx+b

Passes through (1,1) and the center of the circle x^2-4x+y^2-8y+16=0
I got it down to find that the center was (2,4) but then I got stuck there.
Please I really need help because I have a big test tomorrow!!
Thanks!
Click to expand...

Equation of a straight-line through (x[sub:3dmhr1ua]1[/sub:3dmhr1ua], y[sub:3dmhr1ua]1[/sub:3dmhr1ua]) and (x[sub:3dmhr1ua]2[/sub:3dmhr1ua], y[sub:3dmhr1ua]2[/sub:3dmhr1ua]) is:

\(\displaystyle \frac{y-y_1}{y_2-y_1} \, = \, \frac{x-x_1}{x_2-x_1}\)
 
Another more common form for the equation of a line, given two points, is called the Two Point Form:

y - y1 = [(y2 - y1)/(x2 - x1)] (x - x1)

Of course, the expression (y2 - y1)/(x2 - x1) is simply the Slope Formula.
 
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