Locus of the middle points of the chords of the circle `x^(2)+y^(2)-2x-6y-10=0`, which passes through origin is

Options

A. `x^(2)+y^(2)-2x-3y=0`
B. `x^(2)+y^(2)-x-3y=0`
C. `x^(2)+y^(2)-3x+y=0`
D. `x^(2)+y^(2)+3x-y=0`

Answer

To find the locus of the midpoints of the chords of the circle given by the equation \(x^2 + y^2 - 2x - 6y - 10 = 0\) that pass through the origin, we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the circle equation in standard form. We start with:

\[ x^2 + y^2 - 2x - 6y - 10 = 0 \]

We can complete the square for both \(x\) and \(y\).

1. For \(x^2 - 2x\), we complete the square:
\[ x^2 - 2x = (x - 1)^2 - 1 \]

2. For \(y^2 - 6y\), we complete the square:
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting these back into the equation gives:

\[ (x - 1)^2 - 1 + (y - 3)^2 - 9 - 10 = 0 \]

This simplifies to:

\[ (x - 1)^2 + (y - 3)^2 - 20 = 0 \]

Thus, the equation of the circle is:

\[ (x - 1)^2 + (y - 3)^2 = 20 \]

### Step 2: Identify the Center and Radius
From the standard form, we can identify the center \(C(1, 3)\) and the radius \(r = \sqrt{20} = 2\sqrt{5}\).

### Step 3: Midpoint of Chord
Let \(P(h, k)\) be the midpoint of the chord that passes through the origin \(O(0, 0)\). The line segment \(OP\) is perpendicular to the chord at point \(P\).

### Step 4: Slopes and Perpendicularity Condition
The slope of line \(CP\) (from center \(C(1, 3)\) to midpoint \(P(h, k)\)) is given by:

\[ \text{slope of } CP = \frac{k - 3}{h - 1} \]

The slope of line \(OP\) (from origin \(O(0, 0)\) to midpoint \(P(h, k)\)) is:

\[ \text{slope of } OP = \frac{k}{h} \]

Since \(OP\) is perpendicular to \(CP\), the product of their slopes must equal \(-1\):

\[ \frac{k - 3}{h - 1} \cdot \frac{k}{h} = -1 \]

### Step 5: Cross-Multiplying and Rearranging
Cross-multiplying gives:

\[ (k - 3)k = -h(h - 1) \]

This simplifies to:

\[ k^2 - 3k + h^2 - h = 0 \]

### Step 6: Substitute \(h\) and \(k\) with \(x\) and \(y\)
Now, we substitute \(h\) with \(x\) and \(k\) with \(y\):

\[ y^2 - 3y + x^2 - x = 0 \]

Rearranging gives:

\[ x^2 + y^2 - x - 3y = 0 \]

### Final Equation
Thus, the locus of the midpoints of the chords of the circle that pass through the origin is:

\[ x^2 + y^2 - x - 3y = 0 \]