Math question that I need help in?

x^2-6x+y^2-8y=0

(x^2-6x+9)+(y^2-8y+16)-9-16=0 complete the square

(x-3)^2 + (y-4)^2 = 25 put in standard form

vertex is (3,4) and radius is 5

I leave the rest to you. Plot it on graph paper (notice it is a 3, 4, 5 triangle) and the rest should be obvious.

Consider the circle x^2+y^2-6x-8y=0 ------------------ 1or

(x-3)^2 + (y-4)^2 = 5^2

1. Find the radius and center of the circle? radius = 5 and center is (3,4)

2. Find an equation of the tangent line to the circle at point (0,0)

Differentiating 1 we get,

2x +2yy' -6 -8y' = 0 ------------------------------------------- 2.

putting x = 0 and y = 0 we have y' = -3/4

or Required equation is y = (-3/4)x or 4y+3x =0 ------------------------------- 2

3.Find the equation of the tangent line to the circle at the point (6,0)

Putting x = 6 and y = 0 in 2, we have

12 -6 -8y' = 0 or y' at (x =6 and y = 0) = 3/4

Required equation is (y-0) =(3/4)(x-6) or 4y -3x +18 = 0 ------------------------------ 3

4. Where do the 2 tangent lines intersect?

8y + 18 = 0 or y = -9/4 and x = 3

So required point is (3, -9/4)