Question

Equation of the tangent at (1,1) to the circle 2x^2+2y^2-2x-5y+3=0

Answer 1

Answer:

Step-by-step explanation:

2x^2+2y^2-2x-5y+3=0

d/dx(2x^2+2y^2-2x-5y+3) = 4x + 4y.dy/dx - 2 - 5.dy/dx = 0

4y.dy/dx - 5.dy/dx = 2 - 4x

dy/dx = (2- 4x)/(4y - 5)

at (1.1) dy/dx = 2-4 / 4-5 = 2

now equation at (1,1) is

y-1 = 2(x-1)

y-1 = 2x - 2

2x - y -1 = 0