# Thread: [SOLVED] equation of a tangent to circle

1. ## [SOLVED] equation of a tangent to circle

last one now honest

a circle has the equation $x^2+y^2-8x-4y=9$

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!

2. Originally Posted by coyoteflare
last one now honest

a circle has the equation $x^2+y^2-8x-4y=9$

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.

$m=\frac{7-^-3}{2-6}=-\frac{5}{2}$

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 $\frac{2}{5}$

Now, all you need to do is determine the equation of the line that passes through A(2, 7) with a slope of $\frac{2}{5}$.

Use $y=mx+c$, x = 2, y = 7, and m = $\frac{2}{5}$

Solve for c

$7=\frac{2}{5}(2)+c$

Can you complete it?

3. Originally Posted by coyoteflare
last one now honest

a circle has the equation $x^2+y^2-8x-4y=9$

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!
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.