1. ## coordinate geometry, circle

Hi there,
I have this : The line y = x + a , where a is a constant is a tangent to the circle (x-4)2 + (y-1)2 = 16. Find two possible values of a.

So am i right doing it like this:

(x-4)2 + (x+a-1)2 = 16
x2 - 8x + 16 + x2 + a2 + 1 + 2xa - 2x - 2a - 16 = 0
2x2 - 10x + a2 + 2xa - 2a +1 = 0
now using b2-4ac:
(-10 + 2a)2 - 4 (2) (a2 - 2a + 1) which gives me
-4 (a2 + 6a - 23)
and that gives me possible values of 23.65 and and 12.34 when i should be getting -8.66 or 2.66.

Can anyone tell me what im doing wrong ?

2. ## Re: coordinate geometry, circle

Originally Posted by AbYz
The line y = x + a , where a is a constant is a tangent to the circle (x-4)2 + (y-1)2 = 16. Find two possible values of a.
You posted this in pre-collage, meaning that calculus is not to be used. Is that correct?
Do you know other methods to find points on the circle where the slope is $1~?$

3. ## Re: coordinate geometry, circle

Originally Posted by AbYz
Hi there,
I have this : The line y = x + a , where a is a constant is a tangent to the circle (x-4)2 + (y-1)2 = 16. Find two possible values of a.

So am i right doing it like this:

(x-4)2 + (x+a-1)2 = 16
x2 - 8x + 16 + x2 + a2 + 1 + 2xa - 2x - 2a - 16 = 0
2x2 - 10x + a2 + 2xa - 2a +1 = 0
now using b2-4ac:
(-10 + 2a)2 - 4 (2) (a2 - 2a + 1) which gives me
-4 (a2 + 6a - 23)=0
You are correct up to here. Seems you have made a mistake using the quad formula. I got -8.66 or 2.66. Try it again from here.

and that gives me possible values of 23.65 and and 12.34 when i should be getting -8.66 or 2.66.

Can anyone tell me what im doing wrong ?
see comment in red

4. ## Re: coordinate geometry, circle

Yes i did stupid mistake with formula..

Originally Posted by Plato
You posted this in pre-collage, meaning that calculus is not to be used. Is that correct?
Do you know other methods to find points on the circle where the slope is $1~?$
Hmm , i could not think of anything there to be honest so after googling i did this ^. Can you share with me this other method ? i can only think of simultaneous equations but im not sure of it as i end up with 3 unknowns ( x , y and a) ?

5. ## Re: coordinate geometry, circle

Originally Posted by AbYz
Yes i did stupid mistake with formula..
Hmm , i could not think of anything there to be honest so after googling i did this ^. Can you share with me this other method ? i can only think of simultaneous equations but im not sure of it as i end up with 3 unknowns ( x , y and a) ?
If $(p,q)$ is a point on the circle $(x-h)^2+(y-k)^2=r^2$, then the line determined by the point and the center has slope $m=\dfrac{q-k}{p-h}$, of course there a special case if $p=h$.
A tangent to the circle at $(p,q)$ has slope $m_{\bot}=-m^{-1}$, because it is perpendicular to the radial segment.

Thus in the given problem, we look for points in the circle at which the line to center has slope is $-1$ because the given tangent has slope $1$

6. ## Re: coordinate geometry, circle

not sure im following, i have
$-1 = \frac{q-1}{p-4}$
$-p+4=q-1\\ \ p=5-q\\ \ q=5-p\\ \ p = 2.5\\ \ q= 2.5\\$

what should i do with this ?

7. ## Re: coordinate geometry, circle

Originally Posted by AbYz
not sure im following, i have
$-1 = \frac{q-1}{p-4}$
$-p+4=q-1\\ \ p=5-q$
$((5-q)-2)^2+(q-1)^2=16^2$ SOLVE for $q$ to for $p$