Question

Find the equation of the lines tangent to the circle at point where x=3 x²+y²-14x+4y+33=0?

Answer 1

`2x - y=6` and `2x + y=2`

Explanation:

I'm going to do this using a trick that lets us do calculus on algebraic curves with only Algebra I. Let

`f(x,y) = x^2 + y^2 - 14x + 4y + 33`

The tangent at `(r,s)` on `f` is the best linear approximation to `f(x,y)=0` near `(r,s).` Let's find it.

`f(x,y)=f(r+(x-r), s+(y-s))`

`f(x,y) = (r+(x-r))^2 + (s+(y-s))^2 - 14(r+(x-r)) + 4(s+(x-s))+33`

We'll do the algebra, our goal being a polynomial in the small differences `x-r` and `y-s.` So we don't expand those differences.

`f(x,y) = r^2+ 2r(x-r) +(x-r)^2 + s^2+2s(y-s)+(y-s)^2 - 14r-14(x-r) + 4s+4(y-s)+33`

`f(x,y) = r^2 + s^2 - 14r + 4s +33 + (2r -14)(x-r) +(2s+4)(y-s) +(x-r)^2+(y-s)^2`

`f(x,y) = f(r,s) + (2r -14)(x-r) +(2s+4)(y-s) +(x-r)^2+(y-s)^2`

Since `(r,s)` is on `f` we know `f(r,s)=0`

`f(x,y) = (2r -14)(x-r) +(2s+4)(y-s) +(x-r)^2+(y-s)^2`

What we've done here is to generate what's called the Taylor expansion of `f` at `(r,s)` using middle school math. Since the differences are small, their squares are smaller still, and we can get our best linear approximation by dropping them and setting the whole thing to zero (because we're approximating `f(x,y)=0`).

`0 = (2r -14)(x-r) +(2s+4)(y-s)`

That's the equation for the tangent at `(r,s).`

The question asks for the equations when `r=3.` The tangent is

`0 = (2(3)-14)(x-3) +(2s+4)(y-s) = -8x + (2s+4)y +24 -s (2s+4)`

`f(r,s)=0` so

`0=3^2 + s^2 - 14(3) + 4s + 33 = s^2 + 4s =s(s+4)`

`s= 0` or `s=-4`

Plugging these in we get two tangent lines:

`0 = -8x + (2(0)+4)y +24 - (0) (2s+4)` and `0 = -8x + (2(-4)+4)y +24 - (-4) (2(-4)+4)`

`-8x +4y+24=0` and `-8 x - 4 y + 8 = 0`

`2x - y=6` and `2x + y=2`

I'm going to post then I'll check these with the Socratic `cancel{text{crasher}}` grapher.

Here's the equation I'm going to feed the grapher to plot all these at once:

`0=(2x - y-6)(2x + y-2)( x^2 + y^2 - 14x + 4y + 33)(x-3)`

graph{0=(2x - y-6)(2x + y-2)( x^2 + y^2 - 14x + 4y + 33)(x-3) [-4.3, 15.815, -6.83, 3.23]}

Not the best graphing program ever, but we see we got the correct solution.

Answer 2

`y=2x-6 and y=-2x+2`

Explanation:

`x^2+y^2-14x+4y+33=0`

Using implicit differentiation.

`2x+2ydy/dx-14+4dy/dx+0=0`

`(2y+4)dy/dx = 14-2x`

`dy/dx = (14-2x)/(2y+2) = (7-x)/(y+2)`

Thus the slope of any point `(x,y)` on this circle will be: `(7-x)/(y+2)`

We are asked to find the tangents where `x=3`

At `x=3`

`-> 3^2+y^2-3xx14+4y+33=0`

`9+y^2-42+4y+33=0`

`y^2+4y=0`

`y(y+4)=0 -> y =0 or -4`

The slope of the tangent at `(3,0) = (7-3)/(0+2) = 2`

`:.` the equation of the tangent is: `y-0 = 2(x-3)`

`y= 2x-6`

The slope of the tangent at `(3,-4) = (7-3)/(-4+2) = -2`

`:.` the equation of the tangent is: `y+4 = -2(x-3)`

`y = -2x+2`

We can see the circle and its tangents at `x=3` in the graphic below.