SOLUTION: what is the equation of the line tangent to the circle x^2+y^2=25 at the point (3,4) ? I don't know how to go about this sum. Please solve and explain thanks

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Question 237326: what is the equation of the line tangent to the circle x^2+y^2=25 at the point (3,4) ?
I don't know how to go about this sum. Please solve and explain
thanks
Found 2 solutions by solver91311, eggsarecool:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The general equation of a circle with center at and radius is



Hence the circle



Must be centered at the origin.

The line tangent to the point is perpendicular to the radius segment with endpoints and

Determine an equation of the line that contains the aforementioned radius segment by using the two point form of the equation of a line:



Where and are the coordinates of the given points.

Solve the resulting equation for to put this equation into slope-intercept form. Then determine the slope of the radius segment by inspection of the coefficient on .

Next use the fact that the slopes of two perpendicular lines are negative reciprocals, that is:



Calculate the negative reciprocal of the slope of the radius segment to get the slope of the tangent.

Then use the point-slope form of the equation of a line with the calculated slope and the point to derive the desired equation:



Where is the calculated slope and is the point

John

Answer by eggsarecool(46) About Me  (Show Source):
You can put this solution on YOUR website!
You need to implicit differentiation for this problem and solve for dy%2Fdx
So we have.
x%5E2%2By%5E2=25
The derivative is 2x%2B2y%2Ady%2Fdx=0 the 2y produces the dy%2Fdx because y is a function so you have to apply the chain rule to it.
So now.
2y%2Ady%2Fdx=-2x
dy%2Fdx=-2x%2F2y
dy%2Fdx=-x%2Fy
And x and y are given by (3,4) giving x=3 and y=4. (x%5B1%5D,y%5B1%5D)
So dy%2Fdx=-%283%2F4%29
Now for the tangent line you have.
y-y%5B1%5D=%28dy%2Fdx%29%28x-x%5B1%5D%29
so we have y-4=-%283%2F4%29%28x-3%29
Giving us y-4=-%283%2F4%29x%2B9%2F4 Add 4 to both sides. 4 can be rewritten as 16%2F4
y=-%283%2F4%29x%2B9%2F4%2B16%2F4
y=-%283%2F4%29x%2B25%2F4
And that is your tangent line.