# Easy pre-calc help needed

#### Ellis

##### Member
Okay everyone, I feel like a real idiot right now. I'm taking an easy pre-calc course this summer and today we had our first test. This was the first question on it. It is not complicated at all but I really can't see what I'm doing wrong.

Given: The equation of a circle: (x^2 + y^2 - x + 3y = 20)

a) Write the equation in standard form

b) Find the center and radius

c) Find the equation of a line tangent to the circle at (-1, 3), express your answer in slope-intercept form

d) The points (5,0) and (-4,-3) are on the circle. Find the equation for the line passing through them. Express your answer in slope-intercept form

e) What conclusion can be drawn from the lines in part c and d?

This last part was what really troubled me. The professor said it was incorrect, which makes me think I did some other stuff wrong. I was really frustrated trying to see what went wrong but I ran out of time still struggling to see it. This question is written all from memory but it should be right. I'm not going to get the test back until monday, but I'm really anxious to see why I failed. So either do the work, then compare, or just look over my work. Either way, I just want to see what I'm doing wrong.

a) x^2 - x + y^2 + 3y = 20
(x - 1)^2 + (y + 3/2)^2 = 20 + 1 + 9/4
(x - 1)^2 + (y + 3/2)^2 = 93/4

b) C: (1, -3/2) R = sqrt(93)/2

c) So find the slope of the radius to the point, then take the negative reciprocal to plug m into point slope equation. So (3 -(-3/2))/(-1 -1) = -9/4. So m = 4/9.

(y - 3) = 4/9(x + 1)
y = 4/9(x) + 4/9 + 3
y = 4/9(x) + 31/9

d) m = (0 -(-3))/(5 -(-4) = 3/9 = 1/3
(y + 0) = 1/3(x - 5)
y = 1/3(x) -5/3

e) Ok wth, seriously. I must have done something wrong. I see no conclusion that can be drawn here. I notice that I had done the equation wrong in part c on the test, but I still see no connection between either set of equations. I set the two equations equal to each other to find the location of the x intercept where they meet then plugged in to find y. I went to ask the teacher if that was on the right track for what he was looking for and he had a weird look on his face and said... "ohhhh noo, just try some more stuff." So I figure the two lines were supposed to be parallel but I couldn't figure out what I was doing wrong. So, what am I not seeing?

Edit: Okay, I think I see now that I didn't complete the square correctly in part a. It should be (x - 1/2)^2. So the equation is wrong and the center of the circle and radius are wrong. So part C is wrong, and so part E should be... what... that they are parallel? damn...

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#### imaghost

##### Member
I don't quite understand what you did with a. Isn't it supposed to be (x + y)^2 - x + 3y? I don't know anything else with the problem. I think you are wrong with this part though.

#### Ellis

##### Member
I think I see it now. Since I did the equation wrong in part a, I got the wrong center and radius, so when I was trying to find the equation in part c it was off slightly. I feel really stupid.

I don't quite understand what you did with a. Isn't it supposed to be (x + y)^2 - x + 3y? I don't know anything else with the problem. I think you are wrong with this part though.
Well the goal is to write the same equation as (x - h)^2 + (y - k)^2 = r^2, so not exactly.

#### imaghost

##### Member
I think I see it now. Since I did the equation wrong in part a, I got the wrong center and radius, so when I was trying to find the equation in part c it was off slightly. I feel really stupid.

I don't quite understand what you did with a. Isn't it supposed to be (x + y)^2 - x + 3y? I don't know anything else with the problem. I think you are wrong with this part though.
Well the goal is to write the same equation as (x - h)^2 + (y - k)^2 = r^2, so not exactly.
yeah... I don't really know... I don't know the equation. I remember the stuff, just not the equations and what to do.

#### moogra

##### Member
(x^2 + y^2 - x + 3y = 20)

c) Find the equation of a line tangent to the circle at (-1, 3), express your answer in slope-intercept form

Take derivative and plug in the given values
2x+2y*y'-1+3y'=0
y'(2y+3)=1-2x
y'=(1-2x)/(2y+3)
y'=(1-2(-1))/(2(3)+3)
y'=3/9=1/3

point-Slope time
given point: (-1, 3)
slope: 1/3
y - 3 = -1/3 (x + 1)

slope-intercept
y = -1/3 x + 8/3

d) The points (5,0) and (-4,3) are on the circle. Find the equation for the line passing through them. Express your answer in slope-intercept form

slope: (3-0)/(-4-5)=3/-9=-1/3
Point-slope time
y = -1/3(x-5)
y= -1/3 x + 5/3

e) You can see that they are parallel from this.

#### Ellis

##### Member
Thank you moogra. Aren't the slopes both positive 1/3 though?

I'm really beating myself up over this. The rest of the test was pretty much prefect. I can't believe I didn't see this small mistake.

Edit: Oh I see. I wrote it down wrong... It was (-4, -3) not (-4, 3)... sorry!!

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