Finding the Equation of the Tangent to a Circle at a Given Point

In this article, we will explore how to find the equation of the tangent to a circle at a given point. We will use a specific circle with the equation x^2y^225 and a point on the circle, A(-4,3). Follow along as we solve the problem using both algebraic and calculus-based methods.

Algebraic Method

To begin, let's rewrite the given equation of the circle in a more familiar form:

[x^4 y^2 25]

We start by subtracting x^2 y^2 25 from x^2 y^2 25:

[x^2 - 4^2 y^2 - 3^2 0]

From this, we get:

[8x - 6y 25 -25 quad text{or} quad 4x - 3y -25]

The slope of the radius at point A(-4,3) is -3/4. Since the tangent is perpendicular to the radius, the slope of the tangent is the negative reciprocal of -3/4, which is 4/3. Using the point-slope form of the equation of a line:

[y - 3 frac{4}{3}(x 4)]

Expanding this, we get:

[3y 4x 16 - 9 quad text{or} quad 4x - 3y 25 0]

Calculus Method

Let's now solve the problem using calculus. The equation of the circle is:

[x^2 y^2 25]

We differentiate both sides with respect to x to find the slope of the tangent line:

[2x y frac{dy}{dx} 0]

Therefore, the slope of the tangent line is:

[frac{dy}{dx} -frac{x}{y}]

At the point A(-4,3), the slope of the tangent is:

[frac{dy}{dx} -frac{-4}{3} frac{4}{3}]

Using the point-slope form of the equation of the line, we get:

[y - 3 frac{4}{3}(x 4)]

Expanding this, we again get:

[3y 4x 16 - 9 quad text{or} quad 4x - 3y 25 0]

Alternative Method Using Geometric Properties

The equation of the circle is:

[x^2 y^2 25]

In this form, the center of the circle is at (0,0) and the radius is 5 units. The equation of the tangent line is:

[y - 3 m(x 4)]

Where m is the slope of the tangent. The perpendicular distance from the center to the tangent line is equal to the radius. Therefore:

[frac{|4m - 3|}{sqrt{m^2 1}} 5]

Squaring both sides, we get:

[(4m - 3)^2 25(m^2 1)]

Expanding and simplifying, we get:

[16m^2 - 24m 9 25m^2 25 quad text{or} quad 9m^2 24m - 16 0]

Using the quadratic formula, we solve for m:

[m frac{-24 pm sqrt{24^2 - 4 cdot 9 cdot (-16)}}{2 cdot 9} frac{-24 pm sqrt{816}}{18} frac{-24 pm 24sqrt{3}}{18} frac{-4 pm 4sqrt{3}}{3}]

Since m 4/3 is the only feasible solution, we substitute it back into the equation of the tangent line:

[y - 3 frac{4}{3}(x 4)]

Expanding this, we again get:

[3y 4x 16 - 9 quad text{or} quad 4x - 3y 25 0]

Thus, the equation of the tangent line at point A(-4,3) is:

[4x - 3y 25 0]

Conclusion

In summary, we have solved the problem of finding the equation of the tangent to the circle at the given point using multiple methods. Whether using algebraic manipulation, calculus, or geometric properties, we consistently arrive at the same result:

[4x - 3y 25 0]

This demonstrates the consistency and reliability of the methods used in solving such problems.