RTCNTC said:Tangents are drawn to the circle x^2 + y^2 = 169 at the points (5, 12) and (5, -12). Find the equations of the tangents.
I need the steps.
I know the circle has a radius of sqrt{169} or 13.
RTCNTC said:The center is (0, 0) and radius 13.
RTCNTC said:The distance from the center to the point of tangency is 13.
The radius from the center to the point of tangency is perpendicular.
MarkFL said:Let's work a more general version of the problem...that is, to find the equation of the line tangent to the circle:
\(\displaystyle (x-h)^2+(y-k)^2=r^2\)
at the point on the circle:
\(\displaystyle \left(x_1,y_1\right)\)
where:
\(\displaystyle y_1\ne k\)
When $y_1=k$, then we know the tangent line is simply $x=x_1$.
The slope of the line will be:
\(\displaystyle m=\frac{h-x_1}{y_1-k}\)
And so, using the point-slope formula, the tangent line is:
\(\displaystyle y=\frac{h-x_1}{y_1-k}\left(x-x_1\right)+y_1\)
Arranged in slope-intercept form, we have:
\(\displaystyle y=\frac{h-x_1}{y_1-k}x+\frac{x_1-h}{y_1-k}x_1+y_1\)
For the given problem, we have:
\(\displaystyle (h,k)=(0,0)\)
And so, our formula reduces to:
\(\displaystyle y=-\frac{x_1}{y_1}x+\frac{x_1}{y_1}x_1+y_1=-\frac{x_1}{y_1}x+\frac{x_1^2+y_1^2}{y_1}=-\frac{x_1}{y_1}x+\frac{r^2}{y_1}\)
The first point is:
\(\displaystyle \left(x_1,y_1\right)=(5,12)\)
And so the tangent line is:
\(\displaystyle y=-\frac{5}{12}x+\frac{169}{12}\)
The second point is:
\(\displaystyle \left(x_1,y_1\right)=(5,-12)\)
And so the tangent line is:
\(\displaystyle y=\frac{5}{12}x-\frac{169}{12}\)
Let's plot the circle and the two lines to verify:
Equations of the tangents are mathematical expressions that represent the slope of a curve at a specific point. They describe the rate of change of a curve at a given point and can be used to find the slope of a line tangent to the curve.
The equation of a tangent line can be found by taking the derivative of the curve at the given point and plugging in the coordinates of the point into the derivative. This will give the slope of the tangent line. Then, using the point-slope form of a line equation, the equation of the tangent line can be written with the slope and point.
The point-slope form of a line equation is y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a known point on the line. This form is useful for finding the equation of a tangent line since the slope and point can be easily determined.
Equations of the tangents have many practical applications, such as in physics, engineering, and economics. They can be used to calculate the velocity of a moving object, the rate of change of a chemical reaction, or the marginal cost of production.
Yes, equations of the tangents can be used for any type of curve, including linear, quadratic, exponential, and trigonometric curves. The only requirement is that the curve is differentiable at the given point, meaning that it has a well-defined derivative at that point.