Finding Tangent Lines: Finding equations at the points (6,0) & (0,0)

general formula of a circle is
(x-a)^2 + (y-b)^2 = r^2
where a and b are x-coordinates of the circle and r = radius

so if you convert your equation to this form you can find centre and radius

can you do this?

you need to complete the square for x^2 - 6x and y^2 - 8y

I'll give you a hand with the x terms, then you can do the y terms by yourself.
We have x^2-6x on the left hand side. We need to get it in the form (x-a)^2, so we just need to factor it. The best way to do this in this example is completing the square, and we get:
(x-3)^2-9
Now if you do something similar for the y terms, you can keep the (x-3)^2 + (y-b)^2 on the left hand side and take all constants over to the right hand side to be your radius. (Note: The (x-3)^2 we have so far represents the x co-ordinate of the centre of the circle, which will be 3).
Please let me know if you need any more help.

What textbook are you using by the way and the edition?

Draw two tangent lines, one starting at point (0,0) and (6,0) that are perpendicular to each other.
TIP: the slope of the tangent line is opposite inverse to the slope from inside the circle.
Good luck

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Thank you.