last one now honest
a circle has the equation $\displaystyle x^2+y^2-8x-4y=9$
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!
last one now honest
a circle has the equation $\displaystyle x^2+y^2-8x-4y=9$
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.
$\displaystyle m=\frac{7-^-3}{2-6}=-\frac{5}{2}$
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 $\displaystyle \frac{2}{5}$
Now, all you need to do is determine the equation of the line that passes through A(2, 7) with a slope of $\displaystyle \frac{2}{5}$.
Use $\displaystyle y=mx+c$, x = 2, y = 7, and m = $\displaystyle \frac{2}{5}$
Solve for c
$\displaystyle 7=\frac{2}{5}(2)+c$
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.