find tangent line to the circle x^2 + y^2=100 at point (-6,8)
Well, it seems you've stumbled upon a curvy question! Let's find the tangent line to the circle x^2 + y^2 = 100 at the point (-6, 8).
To find the equation of the tangent line, we need to find the slope first. The slope of a circle at a particular point is given by the negative reciprocal of the derivative of the circle's equation.
Thus, let me put on my mathematical clown hat and differentiate the equation with respect to x:
d/dx (x^2 + y^2) = d/dx (100)
2x + 2y * dy/dx = 0
Now let's find the slope at point (-6, 8). Substituting the values, we have:
2 * (-6) + 2 (8) * dy/dx = 0
-12 + 16 * dy/dx = 0
16 * dy/dx = 12
dy/dx = 12/16
dy/dx = 3/4
So, we have the slope of the tangent line at (-6, 8) is 3/4.
Now it's time to put our mathematical clown shoes on and use the point-slope form of a line to find the equation of the tangent line.
y - y1 = m(x - x1)
y - 8 = (3/4)(x - (-6))
Simplifying this equation, we have:
y - 8 = (3/4)(x + 6)
4y - 32 = 3x + 18
3x - 4y = -50
And there you have it! The equation of the tangent line to the circle x^2 + y^2 = 100 at the point (-6, 8) is 3x - 4y = -50. But please don't expect me to juggle circles for you – mathematical clowns have their limits!
To find the tangent line to the circle at the given point, we need to determine the slope of the tangent line and then use the point-slope form of a line.
Step 1: Find the derivative of the equation of the circle.
The equation of the circle is x^2 + y^2 = 100. To find the derivative, we'll implicitly differentiate the equation with respect to x.
Differentiating both sides of the equation gives us:
2x + 2y * dy/dx = 0
Step 2: Solve for dy/dx.
To find dy/dx, we rearrange the above equation:
2y * dy/dx = -2x
dy/dx = -2x / 2y = -x / y
Step 3: Evaluate the derivative at the given point.
Plug in the x and y values of the given point (-6, 8) into the derivative:
dy/dx = -(-6) / 8 = 6 / 8 = 3/4
Step 4: Write the equation in point-slope form.
Using the point-slope form of a line, we can write the equation of the tangent line:
y - y1 = m(x - x1),
where (x1, y1) is the given point on the circle and m is the slope we found in step 3.
Plugging in the values:
y - 8 = (3/4)(x - (-6))
Step 5: Simplify the equation.
y - 8 = (3/4)(x + 6)
y - 8 = (3/4)x + (3/4)*6
y - 8 = (3/4)x + (9/2)
y = (3/4)x + (9/2) + 8
y = (3/4)x + (9/2) + 16/2
y = (3/4)x + (25/2)
Therefore, the equation of the tangent line to the circle x^2 + y^2 = 100 at the point (-6, 8) is y = (3/4)x + 25/2.
To find the tangent line to a circle at a given point, we can follow these steps:
1. Differentiate the equation of the circle to find the equation of the derivative (slope) at any point on the circle.
2. Substitute the coordinates of the given point into the derivative equation to find the slope at that specific point.
3. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) are the coordinates of the given point, and m is the slope we obtained in step 2.
4. Simplify the equation from step 3 into slope-intercept form, y = mx + b, where b is the y-intercept.
With these steps, let's proceed with finding the tangent line to the circle x^2 + y^2 = 100 at the point (-6, 8):
Step 1: Differentiate the equation of the circle.
To differentiate, we'll treat x as the independent variable and y as the dependent variable:
d/dx (x^2 + y^2) = d/dx (100)
2x + 2y(dy/dx) = 0
Step 2: Substitute the coordinates of the given point.
Since the given point is (-6, 8), let's substitute x = -6 and y = 8:
2(-6) + 2(8)(dy/dx) = 0
-12 + 16(dy/dx) = 0
16(dy/dx) = 12
(dy/dx) = 12/16
(dy/dx) = 3/4
Step 3: Use point-slope form.
Using the coordinates (-6, 8) and the slope (dy/dx = 3/4), we can write the equation of the tangent line:
y - y1 = m(x - x1)
y - 8 = (3/4)(x - (-6))
Step 4: Simplify into slope-intercept form.
y - 8 = (3/4)(x + 6)
y - 8 = (3/4)x + (3/4)(6)
y - 8 = (3/4)x + 9/2
y = (3/4)x + 9/2 + 8
y = (3/4)x + 25/2
Therefore, the tangent line to the circle x^2 + y^2 = 100 at the point (-6, 8) is y = (3/4)x + 25/2.
using calculus,
2x + 2y dy/dx = 0
dy/dx = -x/y
at (-6,8) , dy/dx = slope = 6/8 = 3/4
non-calculus:
slope of radius from (0,0) to (-6,8)
= (8-0)/(-6-0) = -4/3
so slope of the the tangent, which has to be perpendicular, is +3/4
tangent equation:
y = (3/4)x + b
8 = (3/4)(-6) + b
b = 8+9/2 = 25/2
tangent: y = (3/4)x + 25/2
or 3x - 4y = -50