Circle x^2 + y^2 = 17 and tangent lines at (-4, 1)

[Guest]

I cant figure this problem out at all, can anyone help me?

the equation of a circle and a point on the circle is given. Write an equation of the line that is tangent to the circle at that point

x^2 + y^2 = 17: (-4, 1)

i don't even know how to start its been too long.

any help much appreciated

 

[stapel]

Hint: Tangent lines are, by definition, perpendicular.

What is the slope of the line through the center and the given point? What then is the perpendicular slope? And so forth.

Eliz.

 

[soroban]

Hello, adon!

The equation of a circle and a point on the circle is given.
Write an equation of the line that is tangent to the circle at that point.
. . \(\displaystyle x^2\,+\,y^2 \:=\: 17,\;(-4, 1)\)

We can differentiate implicitly to find the slope.

. . \(\displaystyle 2x\,+\,2y\cdot\frac{dy}{dx}\:=\:0\;\;\Rightarrow\;\;\frac{dy}{dx}\:=\:-\frac{x}{y}\)

At \(\displaystyle (-4,1),\;\frac{dy}{dx} \:=\:-\frac{-4}{1} \:=\:4\)


We have a point on the tangent \(\displaystyle (-4,1)\) and the slope of the tangent: \(\displaystyle 4\)

You can now write the equation of the tangent . . .

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Eliz. Stapel has the cleverest idea ... no Calculus needed.

She meant to say:
. . A radius and a tangent to a common point are perpendicular.

The radius from the center \(\displaystyle (0,0)\) to \(\displaystyle (-4,1)\) has slope: \(\displaystyle \frac{1\,-\,0}{-4\,-\,0} \:=\:-\frac{1}{4}\)

So the tangent has slope \(\displaystyle +4\) . . . see?