The circle shown below has equation (x-6)^2+(y+2)^2=58

Question

(a) Verify the point (3,5) is on the graph of the circle

(b) Find the equation of the line tangent to the circle at the point (3,5)

[Recall: A line tangent to a circle is always perpendicular to the radius drawn to the point of tangency.]

Answer 1

a) Plug in x =3 and y = 5

(-3)^2 + 7*2 = 58? True

b) Slope of radius passing through (3, 5) is (-2-5)/)(6 - 3) = -7/3 so,slope of tangent is m = -(-3)/7 = 3/7

y-5 = m{x-3)

y-5 = (3/7)(x-2)