A=(9,0) find the tangent to the circle (x-2)^2+(y-1)^2=50 at A

A=(9,0) find the tangent to the circle (x-2)^2+(y-1)^2=50 at A,

please explain in steps how to answer this question

thank you

(x-2)^2+(y-1)^2=50

differentiate both sides with respect to x
2(x-2) +2(y-1) dy/dx = 0
2(y-1) dy/dx = -2(x-2)
dy/dx = (2-x)/(y-1)

dy/dx = (2-x)/(y-1)
At A(9,0), dy/dx = (2-9)/(0-1) = 7

Equation of tangent at (9,0) is:
y-0 = 7(x-9)
y=7x-63

Differentiate implicity.

2(x-2) +2(y-1) dy/dx =0


dy/dx = [ -2(x-2) ] / [2(y-1)] = -(x-2) / (y-1)

At (9,0), dy/dx = -(7) / -1 = 7

y-0=7(x-9)

y=7x-63

Differentiating both sides:

2(x - 2) + 2(y - 1)*y' = 0

Solve for y' and plug in 9 to find the slope of the tangent. Then find your tangent eqn.