precal help- circles?

y = 8x, i.e.
8x - y = 0

Radius of the circle
= Distance between the center and the tangent
= |[8(3) - (-4)] / √(8² + 1²)|0
= 28/√65

Equation for the circle:
(x - 3)² + (y + 4)² = (28/√65)²
65(x - 3)² + 65(y + 4)² = 784

The line from the center of the circle to the point of tangency will be perpendicular to the line y=8x.
The line will have slope -1/8 and go through (3, -4).
Therefore it has equation
y+4 = -(x-3)/8 =-8x +3/8

Solve for the point of intersection by substituting y = 8x
y+4 = -y + 3/8
2y = -29/8
y = -29/16
-29/16 = 8x
x = -29/128

Now find r by getting the distance between (3, -4) and (-29/128, 29/16)
Then the equation is
(x-3)^2 + (y+4)^2 = r^2

A circle with center (3, -4) would have this general equation for some constant k:

x² + y² - 6x + 8y + k = 0

Substitute y = 8x.

x² + (8x)² - 6x + 8(8x) + k = 0
65x² + 58x + k = 0

The curves are tangent, so the discriminant is zero.

(58)² - 4(65)(k) = 0
3364 - 260k = 0
k = 841/65

x² + y² - 6x + 8y + 841/65 = 0
65x² + 65y² - 390x + 520y + 841 = 0