# Circle and tangient question, AS core 2

1. May 21, 2006

### Sink41

Doing a core 2 maths question, realised i cant remember how to find where a tangient and circle meet.

The circle equation provided in question was x^2 + y^2 -10x + 9 = 0

same as (x - 5)^2 + y^2 = 4^2

Question was:

"Given that line l with gradient 7/2 is a tangient to the circle, and that l touches circle at point T

find an equation that passes through the centre of the circle and T"

i tried to find out where line and circle met but wasnt able too. In mark scheme they had a very easy way to do it (m1m2=-1 so gradient is -2/7 and you know the co-ordinates of the centre of the circle, so you use y-y1 = m(x-x1) )

So, anyway i tried to put line and circle together and realised i couldnt... this is what i did how do you do it?

What i first was say that the forumula of the straight line is 7/2x + c = y where c is a constant

i then substituted that in the circle forumula to get (x-5)^2 + (7/2x + c)^2 = 4^2

multiplied out to get (53/4)x^2 + (7c - 10)x + 9 + c^2 = 0

since there can only be one result, b^2 - 4ac = 0 so

(7c -10)^2 - 4 * (53/4) * (9 + c^2) = 0

which ends up with

102c^2 -140c + 577 = 0

which does not have a result...

2. May 21, 2006

### Hootenanny

Staff Emeritus
HINT: The radius of a circle is always perpendicular to the tangent. What can you say about the gradient two perpendicular lines?

~H

3. May 21, 2006

### Sink41

Thats the method they used in the mark scheme:

so gradient is -2/7, midpoint of circle is 5,0 and you get y = -2/7(x-5)

But i realised that i didnt know how to get the equation of the original line or where it meets the circle, so i thought i should ask about that here (i really shouldnt have confused it by keeping original question in my post)

Last edited: May 21, 2006
4. May 21, 2006

### Hootenanny

Staff Emeritus
Sorry, a missunderstanding on my part. Okay, you now have the equation of the line by using, y - y1 = m(x - x1). Now, if a line and a curve intersect their x and y co-ordinates must be equal at that point. Can you go from here?

~H

5. May 21, 2006

### Sink41

ahhh i get it now you do the line going through the centre of the circle... wont bother posting calculation but:

x = 5 +/- (784/53)^0.5

thanks for the help!

Last edited: May 21, 2006