The mid point of chord by the circle `x ^(2) + y ^(2) = 16 ` on the line `x + y + 1=0` is

Options

A. `((1)/(2), (1)/(2))`
B. `(- (1)/(2) , - (1)/(2))`
C. `((1)/(2) , (-3)/(2))`
D. `((3)/(4), - (7)/(4))`

Answer

To find the midpoint of the chord of the circle \( x^2 + y^2 = 16 \) that lies on the line \( x + y + 1 = 0 \), we can follow these steps:

### Step 1: Identify the Circle and Line
The given circle is \( x^2 + y^2 = 16 \), which has a center at \( (0, 0) \) and a radius of \( 4 \) (since \( r^2 = 16 \)). The line equation is \( x + y + 1 = 0 \).

### Step 2: Rewrite the Line Equation
We can rewrite the line equation in slope-intercept form:
\[ y = -x - 1 \]

### Step 3: Find the Slope of the Perpendicular Line
The slope of the line \( y = -x - 1 \) is \( -1 \). The slope of a line perpendicular to this will be the negative reciprocal, which is \( 1 \).

### Step 4: Write the Equation of the Perpendicular Line
Since the perpendicular line passes through the center of the circle \( (0, 0) \) and has a slope of \( 1 \), its equation can be expressed as:
\[ y - 0 = 1(x - 0) \implies y = x \]

### Step 5: Solve the System of Equations
Now, we need to find the intersection of the two lines:
1. \( y = -x - 1 \)
2. \( y = x \)

Substituting \( y = x \) into the first equation:
\[ x = -x - 1 \]
Adding \( x \) to both sides:
\[ 2x = -1 \implies x = -\frac{1}{2} \]
Now substituting \( x = -\frac{1}{2} \) back into \( y = x \):
\[ y = -\frac{1}{2} \]

### Step 6: Conclusion
The coordinates of the midpoint of the chord are:
\[ \left(-\frac{1}{2}, -\frac{1}{2}\right) \]

### Summary
The midpoint of the chord of the circle \( x^2 + y^2 = 16 \) on the line \( x + y + 1 = 0 \) is \( \left(-\frac{1}{2}, -\frac{1}{2}\right) \).

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