Question

The normal drawn at \( (1,1) \) to the circle \( x^2 + y^2 - 4x + 6y - 4 = 0 \) is:

• A. \( 4x + 3y = 7 \)
• B. \( 4x + y = 5 \)
• C. \( 4x + y = 2 \)
• D. \( 4x - y = 3 \)

Hint:

The normal to a circle at a point passes through the center and follows the slope: \[ m = \frac{y_1 - k}{x_1 - h}. \]

Answer 1

The Correct Option is B

Step 1: Convert the Given Circle Equation to Standard Form The given circle equation is: \[ x^2 + y^2 - 4x + 6y - 4 = 0. \] Rearrange the terms: \[ (x^2 - 4x) + (y^2 + 6y) = 4. \] Step 2: Complete the Square
Completing the square for \( x \): \[ (x^2 - 4x) = (x - 2)^2 - 4. \] Completing the square for \( y \): \[ (y^2 + 6y) = (y + 3)^2 - 9. \] \[ (x - 2)^2 - 4 + (y + 3)^2 - 9 = 4. \] \[ (x - 2)^2 + (y + 3)^2 = 17. \] Thus, the equation represents a circle centered at \( (2,-3) \) with radius \( \sqrt{17} \).
Step 3: Find the Equation of the Normal at \( (1,1) \)
The normal at any point on a circle passes through the center \( (h, k) \) and the point of contact \( (x_1, y_1) \).
The slope of the normal is: \[ m = \frac{y_1 - k}{x_1 - h} = \frac{1 - (-3)}{1 - 2} = \frac{4}{-1} = -4. \] The equation of the normal using the point-slope form: \[ y - y_1 = m(x - x_1). \] \[ y - 1 = -4(x - 1). \] \[ y - 1 = -4x + 4. \] \[ 4x + y = 5. \] Step 4: Conclusion
Thus, the correct answer is \( \mathbf{4x + y = 5} \).