Math Problem Statement

Find an equation for the line tangent to the circle x^2 + y^2 = 25 at the point (3, -4). At what other point on the circle will a tangent line be parallel to the tangent line at (3, -4)?

Solution

Let's start by analyzing and solving the problem.

The given circle is: x2+y2=25x^2 + y^2 = 25 which is a circle centered at the origin (0,0)(0, 0) with radius 5, since 25=5\sqrt{25} = 5.

Part (a): Find the equation of the tangent line to the circle at the point (3,4)(3, -4).

The slope of the radius at the point (3,4)(3, -4) is the slope from the origin to this point, which is: slope of radius=y2y1x2x1=4030=43\text{slope of radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 0}{3 - 0} = \frac{-4}{3} Since the tangent line is perpendicular to the radius, the slope of the tangent line is the negative reciprocal of 43-\frac{4}{3}, which is: slope of tangent line=34\text{slope of tangent line} = \frac{3}{4} Using the point-slope form of a line equation, yy1=m(xx1)y - y_1 = m(x - x_1), with slope 34\frac{3}{4} and point (3,4)(3, -4), we get: y(4)=34(x3)y - (-4) = \frac{3}{4}(x - 3) Simplifying: y+4=34(x3)y + 4 = \frac{3}{4}(x - 3) y=34x944y = \frac{3}{4}x - \frac{9}{4} - 4 y=34x254y = \frac{3}{4}x - \frac{25}{4} Thus, the equation of the tangent line is: y=34x254y = \frac{3}{4}x - \frac{25}{4}

Part (b): Find the other point on the circle where a tangent line is parallel to the tangent line in part (a).

For the tangent line to be parallel, it must have the same slope, 34\frac{3}{4}.

We know the equation of the circle is x2+y2=25x^2 + y^2 = 25. The slope of the tangent line at any point on the circle is the negative reciprocal of the slope of the radius at that point. The slope of the radius at any point (x,y)(x, y) is yx\frac{y}{x}, so the slope of the tangent line is xy-\frac{x}{y}.

To find the point where the slope of the tangent line is 34\frac{3}{4}, we set xy=34-\frac{x}{y} = \frac{3}{4}, which gives: xy=34-\frac{x}{y} = \frac{3}{4} x=34yx = -\frac{3}{4}y Now, substitute this into the equation of the circle x2+y2=25x^2 + y^2 = 25: (34y)2+y2=25\left(-\frac{3}{4}y\right)^2 + y^2 = 25 916y2+y2=25\frac{9}{16}y^2 + y^2 = 25 916y2+1616y2=25\frac{9}{16}y^2 + \frac{16}{16}y^2 = 25 2516y2=25\frac{25}{16}y^2 = 25 y2=16y^2 = 16 y=±4y = \pm 4 If y=4y = 4, then x=34(4)=3x = -\frac{3}{4}(4) = -3. So one point is (3,4)(-3, 4).

Thus, the other point where the tangent line is parallel to the one found in part (a) is (3,4)(-3, 4).


Would you like any further details or clarifications? Here are some related questions:

  1. How can we find the equation of a tangent line to any circle?
  2. Why is the slope of the tangent line the negative reciprocal of the slope of the radius?
  3. How does the geometry of a circle influence the relationship between tangent lines and radii?
  4. What are other methods to find the points of tangency on a circle?
  5. How would the solution change if the center of the circle was not at the origin?

Tip: For any circle, the slope of the tangent line at a point can be derived directly from the negative reciprocal of the radius's slope at that point.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
Circles
Tangents
Slopes

Formulas

Equation of a circle: x^2 + y^2 = r^2
Slope of a line: (y2 - y1) / (x2 - x1)
Equation of a line: y - y1 = m(x - x1)

Theorems

The tangent to a circle is perpendicular to the radius at the point of tangency.

Suitable Grade Level

Grades 10-12