Circle and tangient question, AS core 2

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The discussion revolves around solving a Core 2 maths problem involving a tangent line and a circle. The circle's equation is transformed into standard form, and the tangent line's gradient is given as 7/2. Participants highlight the method of finding the tangent point using the relationship between the gradients of perpendicular lines, leading to the conclusion that the gradient of the radius is -2/7. The correct approach involves using the coordinates of the circle's center and the tangent point to derive the equation of the line, ultimately leading to the intersection points. The conversation concludes with the participant successfully finding the intersection points after clarifying the method.
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Doing a core 2 maths question, realized i can't 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 realized 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... :confused:
 
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HINT: The radius of a circle is always perpendicular to the tangent. What can you say about the gradient two perpendicular lines?

~H
 
Hootenanny said:
HINT: The radius of a circle is always perpendicular to the tangent. What can you say about the gradient two perpendicular lines?

~H
Thats the method they used in the mark scheme:

Sink41 said:
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 gradient is -2/7, midpoint of circle is 5,0 and you get y = -2/7(x-5)
But i realized 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 shouldn't have confused it by keeping original question in my post)
 
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Sink41 said:
But i realized 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 shouldn't have confused it by keeping original question in my post)

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
 
ahhh i get it now you do the line going through the centre of the circle... won't bother posting calculation but:

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

thanks for the help!
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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