The intercept cut on y-axis by the circle whose diameter is the line joining the points (-4,3) and (12,-1) is

Options

A. `2sqrt(13)`
B. `4sqrt(13)`
C. `6sqrt(13)`
D. none

Answer

To find the intercept cut on the y-axis by the circle whose diameter is the line joining the points (-4, 3) and (12, -1), we can follow these steps:

### Step 1: Find the center and radius of the circle
The center of the circle can be found by calculating the midpoint of the diameter, which is the line segment joining the points (-4, 3) and (12, -1).

**Midpoint formula:**
\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the values:
\[ \text{Midpoint} = \left( \frac{-4 + 12}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{8}{2}, \frac{2}{2} \right) = (4, 1) \]

### Step 2: Calculate the radius
The radius can be calculated as half the distance between the two endpoints of the diameter.

**Distance formula:**
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Calculating the distance between (-4, 3) and (12, -1):
\[ \text{Distance} = \sqrt{(12 - (-4))^2 + (-1 - 3)^2} = \sqrt{(12 + 4)^2 + (-4)^2} = \sqrt{16^2 + 4^2} = \sqrt{256 + 16} = \sqrt{272} = 4\sqrt{17} \]
Thus, the radius \( r \) is:
\[ r = \frac{4\sqrt{17}}{2} = 2\sqrt{17} \]

### Step 3: Write the equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 4\), \(k = 1\), and \(r = 2\sqrt{17}\):
\[ (x - 4)^2 + (y - 1)^2 = (2\sqrt{17})^2 \]
This simplifies to:
\[ (x - 4)^2 + (y - 1)^2 = 68 \]

### Step 4: Expand the equation
Expanding the equation:
\[ (x^2 - 8x + 16) + (y^2 - 2y + 1) = 68 \]
Combining like terms:
\[ x^2 + y^2 - 8x - 2y + 17 = 68 \]
Rearranging gives:
\[ x^2 + y^2 - 8x - 2y - 51 = 0 \]

### Step 5: Find the intercept on the y-axis
To find the y-intercept, we set \(x = 0\) in the equation:
\[ 0^2 + y^2 - 8(0) - 2y - 51 = 0 \]
This simplifies to:
\[ y^2 - 2y - 51 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = -2\), and \(c = -51\):
\[ y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-51)}}{2(1)} = \frac{2 \pm \sqrt{4 + 204}}{2} = \frac{2 \pm \sqrt{208}}{2} = \frac{2 \pm 4\sqrt{13}}{2} = 1 \pm 2\sqrt{13} \]
Thus, the y-intercepts are:
\[ y = 1 + 2\sqrt{13} \quad \text{and} \quad y = 1 - 2\sqrt{13} \]

### Final Answer
The intercept cut on the y-axis by the circle is \(1 + 2\sqrt{13}\) and \(1 - 2\sqrt{13}\).