Equation of a line tangent to the circle?

Question

Consider the circle of radius 5 centered at (0,0). Find an equation of the line tangent to the circle at the point (3,4).

Please explain this problem to me step by step. I haven't entered calculus yet, so I would know nothing about any calculus concept.

Answer 1

In order to find the equation of a line, you need the slope and a point that you know is on the line.

The slope is easy:  a tangent to a circle is perpendicular to the radius at the point where the line will be tangent to the circle.  In other words, the radius of your circle starts at (0,0) and goes to (3,4).  This slope is equal to rise/run = (y2-y1)/(x2-x1) = (4-0)/(3-0) = 4/3.  A line that is perpendicular to this radius line will have a slope that is the negative reciprocal of the slope you know.  In other words, the line we want will have a slope of -3/4.

The point we know is on the line is (3,4).

So the generic equation of the line is:

(y-y0) = m(x-x0), where

(x0, y0) is the point we know is on the line, or (3,4) in this case

m is the slope, -3/4 in this case

x,y remain as variables

(y - 4) = (-3/4)(x - 3)

y - 4 = (-3/4)x + 9/4

y = (-3/4)x + 25/4

Answer 2

x² + y² = 25

Let point on circumference be A (3,4)

Centre O

m OA = 4/3

Gradient of tangent at (3,4) = - 3/4

Equation of tangent at (3,4) is :-

y - 4 = (- 3/4) (x - 3)

y = (- 3/4) x + 9/4 + 4

y = (- 3/4) x + 25/4

Answer 3

A radial line from (0,0) to (3,4) has slope 4/3.

The tangent line is perpendicular to this radius so its slope is -3/4.

You have the slope and a point on the line, which is enough for an equation of the tangent line:

y-4 = (-3/4)(x-3)  

http://www.flickr.com/photos/dwread/2788...

Answer 4

to find the equation of the tangent line you need slope and a point

you have the point(3,4) but slope in this case can be found through algebra but if the circle's centre is not at the origin then the slope can be only found by differentiation. as you have said that you don't know calculus, then how can you understand the solution because first you need to know the rules i.e(power, chain, product and quotient) and then implicit differentiation. i cannot tell all these things in the solution. it is just the differentiation part that is the problem for you. basically derivative of an equation is its rate of change or its slope

first you need to figure out the equation of the circle

a circle with centre at the origin has the equation

(x^2)+(y^2)=radius^2

so the equation is x^2 +y^2=25

now differentiate with respect to x. you have to use implicit differentiation.

2x+2y(dy/dx)=0

2y(dy/dx)=-2x

dy/dx=-2x/2y

dy/dx=-x/y

dy/dx at (3,4) =-3/4

so the slope is -3/4 and point is also (3,4)

y-4=(-3/4)(x-3)

4y-16=-3x+9

3x+4y-25=0

so the equation is  3x+4y-25=0

Answer 5

makes the formula x^2+y^2=25, right? take derivative and plug in point to find slope

2x+2y(dy/dx)=0

dy/dx=-2x/2y

dy/dx=-x/y

dy/dx=-3/4

then plug into y=mx+b with the point and solve for b.

4=-3/4(3)+b

b=25/4

y=-3/4x+25/4

make it a good day

Answer 6

The equation of the circle with center (0, 0) and radius r is

x² + y² = r²

The equation of tangent to this circle at the point (x', y') on the circle is

x'x + y'y = r²

Thus, equation of tangent to the circle having center (0, 0) and radius 5 at the point (3, 4) on the circle is

3x + 4y = 25.

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Derivation of the formula for tangent without using calculus :

A normal to the circle at the point (x', y') on the circle passes through the center (0, 0) of the circle.

The slope of the normal is (y' - 0) / (x' - 0) = y'/x'

=> slope of the tangent is - x'/y' and as it passes through (x', y'), its equation is

y - y' = - (x'/y') (x - x')

=> x'x + y'y = x'² + y'² = 1 (because (x', y') is on the circle)

Answer 7

here it goes.

find the equation of the line that crosses the origin and the point (3,4). it should be Y=(4/3)X. the line perpendicular to that line is also tangent to that point since that is the only point that the circle and that line intersect. take the opp reciprocal of the line to get the slope of the perpendicular line. so slope m= -3/4. use the point slope formula to get the equation of the line. y – y1 = m(x – x1). Y1=4 and X1=3, m is the slope -3/4. solve for y to get it in slope intercept form. good luck!

Answer 8

Calculus isn't needed here, just remembering some properties of circles.

The line tangent to a circle is also perpendicular to the radius drawn to the point of tangency. So finding the slope of the radius out to the tangent line is easy in this case, since you're starting at the origin and going out to (3.4): rise/run, or 4/3.

The tangent line will be perpendicular to the radius, so it's slope will be a negative reciprocal: -3/4

Now use y = mx + b, substitute in -3/4 for m, and (3,4) for x and y to find b:

4 = -3/4(3) + b

4 = -9/4 + b

b = 25/4

So the equ'n of the tangent line is y = -(3/4)x + 25/4.

Answer 9

Draw a picture (THIS ALWAYS HELPS).  The tangent just touches the circle at 3,4 and only at this point.  You can verify that the equation (y=Mx+B) has a slope M of -4/3.  You can find the constant B from tossing (3,4) into the equation and solving.