Problem 97 - College Algebra

Question

A tangent line to a circle is a line that intersects the circle at exactly one point. The tangent line is perpendicular to the radius of the circle at this point of contact. Write an equation in point-slope form for the line tangent to the circle whose equation is \(x^{2}+y^{2}=25\) at the point \((3,-4)\)

Step-by-Step Solution

Answer

The equation of the tangent line to the circle \(x^{2}+y^{2}=25\) at the point (3,-4) in point-slope form is \(y + 4 = \frac{3}{4} (x - 3)\).

Step 1: Understand the circle and find slope of radius

The equation of the circle is \(x^{2}+y^{2}=25\). The given point (\(3,-4\)) lies on the circle, and it's the point at which the line is tangent to the circle. The radius from the center to this point has a slope that can be calculated using the formula \(slope = y/x = -4/3\). This is because the radius passes through the center of the circle which is at the origin (0,0) and the point (3,-4).

Step 2: Find the slope of the tangent line

As the tangent line is perpendicular to the radius at the point of contact, the slope of the tangent line is the negative reciprocal of the slope of the radius. If the slope of the radius is -4/3, then the slope of the tangent line will be reciprocal of -4/3, which is \(\frac{3}{4}\).

Step 3: Write equation in point-slope form

The point-slope form of a line is \(y - y1 = m(x - x1)\), where \(m\) is the slope and \((x1, y1)\) is the given point on the line. In this problem, the slope (\(m\)) of the tangent line is \(\frac{3}{4}\), and the point on the line is (3,-4). Substituting these values into the equation looks like this: \(y - (-4) = \frac{3}{4} (x - 3)\) which simplifies to \(y + 4 = \frac{3}{4} (x -3)\).