Question

How do you write an equation of the line tangent to `(x-1)^2+(y-1)^2=25` at the point (4,-3)?

Answer 1

`y = 3/4x - 6`

Explanation:

Expand the equation of the circle.

`x^2 - 2x + 1 + y^2 - 2y + 1 = 25`

`x^2 + y^2 - 2x - 2y + 2 = 25`

Differentiate both sides with respect to x using implicit differentiation and the power rule.

`d/dx(x^2 + y^2 - 2x - 2y + 2) = d/dx(25)`

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

`2y(dy/dx) - 2(dy/dx) = 2 - 2x`

`dy/dx(2y - 2) = 2 - 2x`

`dy/dx = (2 - 2x)/(2y - 2)`

`dy/dx = (2(1 - x))/(2(y - 1))`

`dy/dx = (1 -x)/(y - 1)`

`dy/dx = -(x - 1)/(y - 1)`

The slope of the tangent is given by evaluating `f(x, y)` inside the derivative.

`m_"tangent" = -(4 - 1)/(-3 - 1)`

`m_"tangent" = -3/(-4)`

`m_"tangent" = 3/4`

The equation of the tangent is therefore:

`y - y_1 = m(x -x_1)`

`y - (-3) = 3/4(x - 4)`

`y + 3 = 3/4x - 3`

`y = 3/4x - 6`

Hopefully this helps!

Answer 2

`y=3/4x-6`

Explanation:

The equation is that of a circle centred at point `O(1,1)`, Let `P` be the point `(4,-3)`. The tangent at `P` is at right angles to the radius at `P`. But the gradient of radius `OP` is `(-3-1)/(4-1)=-4/3`. Therefore the gradient of the tangent is `(-1)/(-4/3)=+3/4`. Therefore the equation of the tangent is `y=3/4 x + c` for some `c`. Since `P` lies on the tangent, `-3=3/4 4+c`. Hence `c=-6`.