What is the radius of the passing through the point (2 ,4) and having centre at the intersection of the line x - y = 4 and 2x + 3y + 7 = 0?

Options

A. 3 units
B. 5 units
C. `3 sqrt3 ` units
D. `5 sqrt2` units

Answer

To find the radius of the circle passing through the point (2, 4) and having its center at the intersection of the lines \(x - y = 4\) and \(2x + 3y + 7 = 0\), we can follow these steps:

### Step 1: Find the intersection point of the two lines

We have two equations:
1. \(x - y = 4\) (Equation 1)
2. \(2x + 3y + 7 = 0\) (Equation 2)

From Equation 1, we can express \(x\) in terms of \(y\):
\[ x = y + 4 \]

Now, substitute this expression for \(x\) into Equation 2:
\[ 2(y + 4) + 3y + 7 = 0 \]

### Step 2: Simplify and solve for \(y\)

Expanding the equation:
\[ 2y + 8 + 3y + 7 = 0 \]
Combine like terms:
\[ 5y + 15 = 0 \]
Now, solve for \(y\):
\[ 5y = -15 \implies y = -3 \]

### Step 3: Find the corresponding \(x\) value

Now that we have \(y = -3\), substitute this back into the expression for \(x\):
\[ x = -3 + 4 = 1 \]

Thus, the center of the circle \(O\) is at the point \((1, -3)\).

### Step 4: Use the distance formula to find the radius

The radius \(R\) of the circle can be calculated using the distance formula between the center \(O(1, -3)\) and the point \(P(2, 4)\):
\[ R = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ R = \sqrt{(2 - 1)^2 + (4 - (-3))^2} \]
\[ R = \sqrt{(1)^2 + (4 + 3)^2} \]
\[ R = \sqrt{1 + 7^2} \]
\[ R = \sqrt{1 + 49} = \sqrt{50} \]
\[ R = \sqrt{25 \times 2} = 5\sqrt{2} \]

### Final Answer

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

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