Finding Tangent Equations to a Circle

This problem asks us to find the equations of the tangent lines to a given circle at specific points defined by an x-coordinate. The circle's equation is $x^2 + y^2 = 9$. This is a standard form equation for a circle centered at the origin $(0,0)$ with radius squared $r^2 = 9$. Thus, the radius $r = 3$.

Determining Points of Tangency on the Circle

We are given that the tangent lines touch the circle where $x = 2$. To find the exact points on the circle where $x=2$, we substitute this value into the circle's equation:

Now, we solve for $y$:

So, there are two points on the circle where $x=2$. These points are $(2, \sqrt{5})$ and $(2, -\sqrt{5})$. These are the points of tangency.

Formula for Tangent Equation

The equation of the tangent to a circle $x^2 + y^2 = r^2$ at a point of tangency $(x_1, y_1)$ on the circle is given by the formula:

Calculating Tangent Equations at Each Point

We will apply the tangent formula to each of the points of tangency we found. Remember $r^2 = 9$.

Tangent at $(2, \sqrt{5})$

Here, $(x_1, y_1) = (2, \sqrt{5})$. Using the formula $xx_1 + yy_1 = r^2$:

This gives us the equation of the first tangent line.

Tangent at $(2, -\sqrt{5})$

Here, $(x_1, y_1) = (2, -\sqrt{5})$. Using the formula $xx_1 + yy_1 = r^2$:

This gives us the equation of the second tangent line.

Summary of Tangent Equations

The equations of the tangents to the circle $x^2 + y^2 = 9$ at $x=2$ are:

• $2x + \sqrt{5}y = 9$
• $2x - \sqrt{5}y = 9$

These equations match one of the provided options.

Revision Table: Circle Tangent Concepts

Additional Information: Other Tangent Methods

While the formula $xx_1 + yy_1 = r^2$ is very convenient for tangents at a point on the circle when the center is at the origin, there are other methods for finding tangent equations:

• Using Calculus: Find the derivative $\frac{dy}{dx}$ from the circle equation to get the slope of the tangent at $(x_1, y_1)$. Then use the point-slope form of a line ($y - y_1 = m(x - x_1)$).
• Using Discriminant: Assume the tangent equation in a general form (e.g., $y = mx + c$ or $x = k$). Substitute this into the circle equation to get a quadratic in one variable. The discriminant of this quadratic must be zero for a tangent line (since there is only one point of intersection). Solve for the unknown parameters (like $m$ and $c$). This method is often more complex.

For a circle centered at $(h, k)$ with equation $(x-h)^2 + (y-k)^2 = r^2$, the tangent at $(x_1, y_1)$ on the circle is $(x-h)(x_1-h) + (y-k)(y_1-k) = r^2$. The method used in the solution is a special case of this formula when $(h, k) = (0, 0)$.