Question

Find the equation of the tangent to the circle x2 + y2 = 109 at point (-10,3)

Answer 1

Answer:

10x - 3y = -109 .

Step-by-step explanation:

x^2 + y^2 = 109

Implicit Differentiation:

2x + 2y.y' = 0

y' = -2x/2y = -x/y

So at the point (-10,3) the gradient of the tangent is -(-10)/3 = 10/3.

Equation of the tangent:

y - y1 = m(x - x1)

y - 3 = 10/3(x + 10)

y - 3 = 10/3 x + 100/3

3y - 9 = 10x + 100

-109 + 3y = 10x

10x - 3y = -109