SOLUTION: What is the equation of a line that is tangent to the circle (x-2)2 + y2= 25 at the point (6,3)?

Algebra ->  Test  -> Lessons -> SOLUTION: What is the equation of a line that is tangent to the circle (x-2)2 + y2= 25 at the point (6,3)?       Log On


   

Question 310821: What is the equation of a line that is tangent to the circle (x-2)2 + y2= 25 at the point (6,3)?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of the tangent to the circle%28x-2%29%5E2%2By%5E2+=+25 at the point (6,3)
Using implicit differentiation, we get:
2%28x-2%29%2A%281%29%2B2y%2A%28dy%2Fdx%29+=+0 Solve for the tangent (slope), dy%2Fdx
dy%2Fdx+=+-%28x-2%29%2Fy and, at the point (6,3) the slope is:
dy%2Fdx+=+-4%2F3 or m+=+-4%2F3 Now find the equation of the line with slope m+=+-4%2F3 and containing the point (6,3).
y+=+mx%2Bb Substitute m+=+-4%2F3, y+=+3 and x+=+6
3+=+-%284%2F3%29%2A6%2Bb Simplify and solve for b.
3+=+-8%2Bb
b+=+11
The final equation, in slope-intercept form is:
highlight%28y+=+-%284%2F3%29x%2B11%29
-----------------------------------
But there is a simpler way to do this!
Find the equation of the line that is perpendicular to the radius of the circle and which contains the point (6,3).
Recall that the tangent to a circle is perpendicular to the radius at the point of tangency.
Find the slope of the radius whose end points are the circle center at (2,0) and the point of tangency at (6,3).
m+=+%283-0%29%2F%286-2%29
m+=+3%2F4
The line that's perpendicular to this has a slope that's the negative reciprocal of3%2F4 and this is-4%2F3, so you can start with...
y+=+-%284%2F3%29x%2Bb Now substitute the x- and y-coordinates of the point of tangency (6,3).
3+=+-%284%2F3%29%2A6%2Bb Simplify and solve for b.
3+=+-8%2Bb so...
b+=+11
The equation is then...
highlight_green%28y+=+-%284%2F3%29x%2B11%29