Tangent Line of a Circle

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October 6th 2009, 06:38 PM
Vesicant

Tangent Line of a Circle

Hey guys, I've tried this one a few times and just can't seem to get it. Help would be appreciated:

A circle has its center at C(4,8). A point Q(12,6) is on the circumference of the circle. A tangent line to the circle is drawn at point Q. Determine the equation of the tangent line.
October 6th 2009, 07:03 PM
mr fantastic

Quote:

Originally Posted by Vesicant

Hey guys, I've tried this one a few times and just can't seem to get it. Help would be appreciated:

A circle has its center at C(4,8). A point Q(12,6) is on the circumference of the circle. A tangent line to the circle is drawn at point Q. Determine the equation of the tangent line.

Since you originally posted this in the Pre-algebra and Algebra subforum (and I moved it to here) I assume that you're to do this without using calculus ..... Correct?
October 6th 2009, 07:08 PM
Vesicant

Quote:

Originally Posted by mr fantastic

Since you originally posted this in the Pre-algebra and Algebra subforum (and I moved it to here) I assume that you're to do this without using calculus ..... Correct?

Correct
October 6th 2009, 08:19 PM
mr fantastic

Quote:

Originally Posted by Vesicant

Correct

Draw a diagram. Note the fact that the angle between the tangent and the radius is 90 degrees. Use this fact, geometry and some simple trigonometry to get the angle $\theta$ that the tangent makes to the x-axis. Then $m = \tan \theta$ .

Knowing the gradient of a line and a point on it, you should be able to get the equation of the line.
October 7th 2009, 12:15 PM
HallsofIvy

You might also use the fact that if the lines corresponding to $y= m_1x+ b_1$ and $y= m_2 x+ b_2$ are perpendicular, then $m_1m_2= -1$ . You know that a radius passing through (4, 8) and (12, 6). What is the slope of that line? You know that the tangent at (12, 6) is perpendicular to that radius so you can find the slope of the tangent. Now you know the slope and one point of the line so you can find the equation of the line.
October 7th 2009, 05:07 PM
Vesicant

Quote:

Originally Posted by HallsofIvy

You might also use the fact that if the lines corresponding to $y= m_1x+ b_1$ and $y= m_2 x+ b_2$ are perpendicular, then $m_1m_2= -1$ . You know that a radius passing through (4, 8) and (12, 6). What is the slope of that line? You know that the tangent at (12, 6) is perpendicular to that radius so you can find the slope of the tangent. Now you know the slope and one point of the line so you can find the equation of the line.

So if the slope given by (4,8) and (12,6) is $M=8-6/4-12=2/-8=-1/4$ then the slope of the tangent is $4/1$ which would make the equation of the tangent $Y=4/1x+3$ ?

Here's a quick graph I did of it, Something doesn't feel right though :/

http://i38.tinypic.com/9rrb7o.jpg
October 8th 2009, 04:06 AM
HallsofIvy

How did you get "+3"? Yes, the slope is 4 so the equation is of the form y= 4x+ b.
In order that it pass through (12, 6) you must have 6= 4(12)+ b so b= 6- 48= -42.
Try y= 4x- 42.
October 8th 2009, 11:14 AM
Vesicant

Quote:

Originally Posted by HallsofIvy

How did you get "+3"? Yes, the slope is 4 so the equation is of the form y= 4x+ b.
In order that it pass through (12, 6) you must have 6= 4(12)+ b so b= 6- 48= -42.
Try y= 4x- 42.

Yeah I know what I did wrong lol, Thought about that while sitting in class today for some reason I put the X intercept into the equation haha.

So yes [Math]Y=4x-42[/tex] is what I got after reworking it, Thanks for all the help.