Finding equation of line?

Question

The point (4, 3) lies on the circle x^2 + y^2 = 25

Determine the slope of the line tangent to the circle at (4, 3) and using the slope find the equation of the tangent line

Answer 1

GIVEN

x^2 + y^2 = 25

take the derivative

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

Solving for dy/dx,

(dy/dx) = -x/y and at point (4,3),

(dy/dx) = -4/3

The slope of the line tangent to the circle at (4,3) is equal to (-4/3).

Therefore, the equation of the tangent line is

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

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

-4x + 16 = 3y - 9

Rearranging,

3y = -4x + 25  ---- this is the equation of the line tangent to the given circle at point (4,3).

Answer 2

2x+2yy'=0

y'=-x/y

y'=-4/3

y=mx+b

m=-4/3

y=3

x=4

3=(-4/3)4+b

32=-16/3+b

96/3+16/3=b

112/3=b

37 and 1/3=b

Answer 3

Whenever you have a problem with a line tangent to a circle, just remember that the tangent line is perpendicular to the radius that's drawn from the center of the circle to the point where they meet.  So, our tangent line is perpendicular to the radius from the center to (4, 3).

Our center is (0,0) since the equation for the circle is x^2 + y^2 = 25.  If our center was not (0,0), we'd have an equation like (x - 3)^2 + (y + 3)^2 = 25.

So, our radius goes from (0, 0) to (4, 3).  The slope of this radius is

(y2 - y1) / (x2 - x1) = (3 - 0) / (4 - 0) = 3 / 4

Now, since our tangent is perpendicular to the radius, the slope of the tangent is the negative reciprocal of the slope of the radius.  The negative reciprocal of 3/4 is -4/3.

Answer:  The slope of the tangent line is -4/3.

Now, to find the equation of the tangent line, we just plug in m = slope = -4/3 and the point (4, 3) into the point-slope equation.

y - y1 = m(x - x1)

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

Distribute

y - 3 = -4/3x + 16/3

Add 3 to both sides

Answer: y = -4/3x + 25/3

Hope this helps you!