Finding the Radius of a Circle

The problem asks us to find the radius of a circle that passes through a specific point (2, 4) and has its center located at the intersection point of two given lines: x – y = 4 and 2x + 3y + 7 = 0.

To solve this, we need to follow these steps:

1. Find the coordinates of the center of the circle by determining the intersection point of the two lines.
2. Calculate the distance between the center and the point (2, 4). This distance will be the radius of the circle.

Step 1: Finding the Center of the Circle

The center of the circle is the solution to the system of linear equations representing the two lines:

Equation 1: \( x - y = 4 \)

Equation 2: \( 2x + 3y + 7 = 0 \)

We can solve this system using the substitution method or elimination method. Let's use substitution.

From Equation 1, we can express \( x \) in terms of \( y \):

\( x = y + 4 \)

Now, substitute this expression for \( x \) into Equation 2:

\( 2(y + 4) + 3y + 7 = 0 \)

Distribute the 2:

\( 2y + 8 + 3y + 7 = 0 \)

Combine like terms:

\( (2y + 3y) + (8 + 7) = 0 \)

\( 5y + 15 = 0 \)

Subtract 15 from both sides:

\( 5y = -15 \)

Divide by 5:

\( y = \frac{-15}{5} \)

\( y = -3 \)

Now substitute the value of \( y \) back into the expression for \( x \):

\( x = y + 4 \)

\( x = -3 + 4 \)

\( x = 1 \)

So, the intersection point of the two lines, which is the center of the circle, is (1, -3).

Let the center of the circle be \( (h, k) = (1, -3) \).

Step 2: Calculating the Radius of the Circle

The radius of the circle is the distance between its center \( (h, k) = (1, -3) \) and the point on the circle \( (x, y) = (2, 4) \).

We use the distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \):

\( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Let \( (x_1, y_1) = (1, -3) \) (the center) and \( (x_2, y_2) = (2, 4) \) (the point on the circle).

The radius \( r \) is:

\( r = \sqrt{(2 - 1)^2 + (4 - (-3))^2} \)

\( r = \sqrt{(1)^2 + (4 + 3)^2} \)

\( r = \sqrt{1^2 + 7^2} \)

\( r = \sqrt{1 + 49} \)

\( r = \sqrt{50} \)

To simplify \( \sqrt{50} \), we find the largest perfect square that divides 50, which is 25:

\( \sqrt{50} = \sqrt{25 \times 2} \)

\( \sqrt{50} = \sqrt{25} \times \sqrt{2} \)

\( \sqrt{50} = 5\sqrt{2} \)

So, the radius of the circle is \( 5\sqrt{2} \) units.

Summary of Radius Calculation

• Center coordinates: \( (1, -3) \)
• Point on circle: \( (2, 4) \)
• Distance Formula: \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
• Calculation: \( r = \sqrt{(2 - 1)^2 + (4 - (-3))^2} = \sqrt{1^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \)

The radius of the circle is \( 5\sqrt{2} \) units.

Revision Table: Circle Geometry Concepts

Additional Information: Solving Systems of Linear Equations

Finding the intersection point of two lines is a fundamental skill in coordinate geometry and algebra. There are several methods to solve a system of two linear equations with two variables (\(x\) and \(y\)):

• Substitution Method: Solve one equation for one variable (e.g., \(x\) in terms of \(y\)), and then substitute that expression into the other equation. This reduces the system to a single equation with one variable.
• Elimination Method: Multiply one or both equations by constants so that the coefficients of one variable are opposites. Add the equations together to eliminate that variable, resulting in a single equation with one variable.
• Graphical Method: Graph both lines on the same coordinate plane. The point where they intersect is the solution. This method is useful for visualization but may not be precise if the intersection coordinates are not integers.

In this problem, we used the substitution method to find the center (1, -3).