find the equation of the tangent to the circle with the equation:( pre calc)

1. x^2 + y^2 +6x-10y+17=0

Using calculus, the tangent at any point on the circle can be found using implicit differentiation....so we have...

2x + 2yy'  + 6 -10y' + 0 = 0

y' [ 2y - 10y ]  =  - [2x + 6 ]

2y' [ y - 5]  =  -2 [ x + 3 ]     divide both sides by 2

y' [ y - 5]  =   - [x + 3]   

y'  = [ x + 3]  / [ 5 - y]

2. .x^2 + y^2 -4x-17=0 at the point (-1,-3) (pre calc)

Notice that the point (-1,-3)  doesn't lie on the graph of this circle : https://www.desmos.com/calculator/z6kse7svjo

????

OK....so we have

2. .x^2 + y^2 -4x-14=0 at the point (-1,-3) (pre calc)

Using implicit differentiation again, the slope at any point is :

2x + 2yy'  - 4 - 0 = 0

2yy' =  4 - 2x    divide through by 2

yy'  = 2 - x      divide by y

y '  = [ 2 - x] / y        so.....at (-1, -3 ), the slope  is     [2 - (-1)] / [ -3]  =  3/-3  = -1

So.......using the point-slope form, we have

y - (-3)  = -1 [ x - (-1) ]

y + 3 = -1 [ x + 1]

y + 3   = -x - 1     subtract 3 from both sides

y = -x - 4

Here's thje graph :  https://www.desmos.com/calculator/m4qjkfosbw

2. .x^2 + y^2 -4x-14=0 at the point (-1,-3) (pre calc)

Here is a non-calculus solution for this one......first.....let's complete the square on x so that we can find the center of the circle.....so we have

x^2 - 4x  + 4  + y^2  =  14  + 4

(x - 2)^2 + ( y - 0)^2   = 18

The center is (2, 0)

Now......a radius drawn from the center to (-1, -3)  will be perpendicular to the tangent line to the circle at (-1, -3).........and the slope of this radial line will be the negative reciprocal of the tangent line.....so the slope of the radial line =   [ -3 - 0] / [ -1 -2]  =  -3/-3   =  1

And the negative reciprocal of this = -1......so this is the slope of the tangent line......and the rest of the problem is worked as before.........

The equation of the tangent to the circle

\(\displaystyle x^{2}+y^{2} - 4x-14=0\)

at the point \((x_{1},y_{1})\)

is given by

\(\displaystyle xx_{1}+yy_{1}-2(x+x_{1})-14=0.\)

The rule is,

replace \(x^{2} \text{ by } xx_{1} ,\)

replace \(y^{2} \text{ by } yy_{1} ,\)

replace \(2x \text{ by } (x + x_{1})\),

and replace (not needed here), \(2y \text{ by } (y + y_{1})\).

Now just substitute (-1, -3) for \((x_{1}, y_{1})\).