Mathematics, 23.02.2020 19:54 codyjacksnow

Find the equation of a line tangent to the circle (x+1)^2+(y−4)^2=25 at the point (3, 1).

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Answer from: gchippewa81

Slope-intercept form: $y=\frac{4}{3}x-3$

Point-slope form: $y-1=\frac{4}{3}(x-3)$

Standard form: 4x-3y=9

Step-by-step explanation:

Calculus Approach:

We will differentiate to find the formula for the slopes of tangent lines around this circle.

(x+1)^2+(y-4)^2=25

Scratch paper:

$\frac{d}{dx}(x+1)^2=2(x+1)\frac{d}(x+1)}{dx}$ by chain rule

$\frac{d}{dx}(x+1)^2=2(x+1)(1+0)$ by power rule and constant rule

$\frac{d}{dx}(x+1)^2=2(x+1)(1)$ by additive inverse property

$\frac{d}{dx}(x+1)^2=2(x+1)$ by multiplicative inverse property

$\frac{d}{dx}(y-4)^2=2(y-4)\frac{d(y-4)}{dx}$ by chain rule

$\frac{d}{dx}(y-4)^2=2(y-4)(y'-0)$ by $\frac{dy}{dx}=y'$ and constant rule

$\frac{d}{dx}(y-4)^2=2(y-4)y'$ by additive inverse property

$\frac{d}{dx}(25)=0$ by constant rule

Back to the problem:

Differentiating (x+1)^2+(y-4)^2=25 gives 2(x+1)+2(y-4)y'=0

Let's solve for .

Subtract 2(x+1) on both sides:

2(y-4)y'=-2(x+1)

Divide both sides by 2:

(y-4)y'=-(x+1)

Divide both sides by (y-4) :

$y'=\frac{-(x+1)}{y-4}$

Now we want to find the slope of the line at (3,1) and we just found our slope formula for the tangents around our circle.

So let's plug it in to find the slope of the tangent line at this point:

$y'=\frac{-(3+1)}{1-4}$

$y'=\frac{-4}{-3}$

$y'=\frac{4}{3}$

So the slope of the line we are looking for is $\frac{4}{3}$ and it goes through (3,1) .

Using point-slope form we have:

$y-1=\frac{4}{3}(x-3)$

Add 1 on both sides:

$y=\frac{4}{3}(x-3)+1$

Distribute:

$y=\frac{4}{3}x-4+1$

Simplify:

$y=\frac{4}{3}x-3$

We can also write in standard form.

Subtract $\frac{4}{3}x$ on both sides:

$\frac{-4}{3}x+y=-3$

Multiply both sides by -3:

4x-3y=9

Geometric/Algebraic Approach:

I provided a rough sketch of what we are looking at.

Tangent lines form a 90 degree angle with the radius at their point of tangency. This means my gray and green line are perpendicular lines. Perpendicular lines have opposite reciprocal slopes (provided the lines we are talking about are not the horizontal/vertical ones).

So let's find the slope of the green line since we know two points on it.

(-1,4)

(3,1)

---------------Subtract and put 2nd number over 1st. This will give us our slope.

-4,3

The slope of the green line is $\frac{3}{-4}$ .

The opposite reciprocal of that will give us the slope of the gray line.

The opposite reciprocal of $\frac{3}{-4}$ is $\frac{4}{3}$ .

Now that we know the slope is $\frac{4}{3}$ and a point, (3,1) , on the line we want to find, we can go on to find the equation. I prefer to just go ahead and use point-slope form again.

$y-1=\frac{4}{3}(x-3)$ .

This is the same equation we obtain the calculus way.

Find the equation of a line tangent to the circle (x+1)^2+(y−4)^2=25 at the point (3, 1).

Answer from: Quest

the correct answer is d

hope this

Answer from: Quest

$answer \:$ paul's with a y-intercept of $ 2856

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