Consider the circle of radius 5 whose center is (0,0). Write the equation of the line tangent to the circle at (3,4).
\[x^2+y^2=r^2\]where r is the radius.
What will be the equation of the circle with a radius 5?
the slope of a tangent to a circle is the perp slope of the line from the center
X^2 + y^2=10?
slope from center to point is:\[\frac{y_1-y_0}{x_1-x_0}\]perp slope is just flip and negate\[-\frac{x_1-x_0}{y_1-y_0}\]
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Ryan is answering a confused reading of your question ....
Well is my method still wrong then?
So the slope would be -3/4
Isolating y and then finding the derivative and then plugging the given point and you find the slope. Then use point slope form with the slope you found and the given point
Hi, suppose you have a circle \[ x^2 + y^2 = r^2 \] then the tangent line to this circle is given as: \[ xx_1 + yy_1 = r^2 \] at any point (x1,y1) on the circle Now substitute x1 = 3 and y1 = 4: \[ 3x + 4y = 5^2 \] \[ 3x + 4y = 25 \]
if your employing calculus, then no
slope of the tangent line is: -3/4 yes
My method is tedious but I still have to use a derivative so is it still wrong?
Ryan, your method reads fine now that i know the method your thinking of :)
Thanks every body I think I got now! :)
something seems wrong with the posting mechanism; anyone else getting multiple posts of the same thing?
Brohoofs for everyone "brohoof" good job guys!
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