Question

The tangent to the circle x^2+y^2=18 is parallel to the tangent x+y=6

Answer 1

Answer:

y = -x - 6

Step-by-step explanation:

If we are looking for the equation of the line tangent to the circle, we have to find the first derivative of the circle function.  If that tangent line is to be parallel to the given tangent line, we need to find the slope of the the given tangent line and make sure the slope of the line we are looking for is the same.  First let's find the slope of the given tangent.  If the given tangent is x+y=6, then to find the slope, solve for y:

y = -x + 6.  So the slope of that line, and also the slope of the tangent we are solving for, is -1.  Hold that thought while we find the derivative of the function.  Using implicit differentiation, we find the derivative to be:

$2x+2y\frac{dy}{dx}=0$

Solving for the slope (dy/dx) gives us:

$2y\frac{dy}{dx}=-2x$ and

$\frac{dy}{dx}=\frac{-2x}{2y}$ so

$\frac{dy}{dx}=-\frac{x}{y}$

Subbing in the value of the slope we found:

$-1=-\frac{x}{y}$ so

-y = -x or, equivalently,

y = x.  Now that we know that y and x are the same value, we can go back to the original circle to find out what they both are by substitution.  If y = x, we make the substution:

If $x^2+y^2=18$, then

$y^2+y^2=18$ and

$2y^2=18$ and

$y^2=9$ so

y = ±3

We will choose the negative root (you'll see why in a second) and sub that in for both y and x since y and x are th same number.  Going back to the fact that the slope is -1:

y - (-3) = -1(x - (-3)) simplifies a bit to:

y + 3 = -x - 3 which gives us, in slope-intercept form:

y = -x - 6

That is the equation of the tangent to the circle that is parallel to the given tangent.

If we would have chosen the principle (or positive) root of 3, our equation would have looked like this:

y - 3 = -1(x - 3) and

y = -x + 3 + 3 so

y = -x + 6.  Notice that that is the EXACT SAME EQUATION as the given tangent.  That's why we have to pick the negative 3 as our root.

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