last one now honest
a circle has the equation
centre is C (4, 2)
AB is diameter A (2,7) and B (6, -3)
find equation of the tangent to the circle at A. give your answer in the form y=mx+c
thanks!
Hello again coyote,
First, find the slope of line BA.
The tangent line (t) through A(2, 7) will be perpendicular to BA.
The slope of the tangent line (t) will be the negative reciprocal of the slope of BA or
Now, all you need to do is determine the equation of the line that passes through A(2, 7) with a slope of .
Use , x = 2, y = 7, and m =
Solve for c
Can you complete it?
The line from (4, 2) to (2, 7), a radius of the circle, has slope (7- 2)/(2- 4)= -5/2. The tangent at that point must be perpendicular to the radius and so must have slope 2/5. The line through (2, 7) with slope 2/5 is y= (2/5)(x- 2)+ 7.
Alternatively, differentiate the equation with respect to x: 2x+ 2y dy/dx- 8- 4 dy/dx= 0 so (2y- 4)dy/dx= 8- 2x. At x= 2, y= 7, that is (14- 4)dy/dx= 8-4 or dy/dx= 4/10= 2/5 just as before. Again, the line through (2, 7) with slope 2/5 is y= (2/5)(x- 2)+ 7.