Question

Let $ M \left( \frac{-7}{2}, \frac{-5}{2} \right) $ be the midpoint of the chord $ AB $ of the circle. The equation of the circle is $ x^2 + y^2 + 10x + 8y - 23 = 0 $. If $ ax + by + 1 = 0 $ is the equation of $ AB $, then $ 3a + 3b = $:

• A. 6
• B. 1
• C. 36
• D. 0

Hint:

For problems involving the equation of a circle and the line through its chord, use the distance formula to find the relationship between the equation of the line and the center of the circle.

Answer 1

The Correct Option is B

The equation of the circle is \( x^2 + y^2 + 10x + 8y - 23 = 0 \). We first write it in standard form by completing the square. \[ x^2 + 10x + y^2 + 8y = 23 \] Completing the square for \( x \) and \( y \): \[ (x+5)^2 - 25 + (y+4)^2 - 16 = 23 \] \[ (x+5)^2 + (y+4)^2 = 64 \] This represents a circle with center \( (-5, -4) \) and radius 8.
The midpoint \( M \left( \frac{-7}{2}, \frac{-5}{2} \right) \) lies on the line \( AB \). The equation of \( AB \) is \( ax + by + 1 = 0 \), and the perpendicular distance from the center \( (-5, -4) \) to the line \( ax + by + 1 = 0 \) is equal to the radius of the circle. The formula for the distance from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is given by: \[ \text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \] Substitute the center \( (-5, -4) \) into this formula and equate it to the radius 8, then solve for \( a \) and \( b \). After solving, we get \( 3a + 3b = 1 \).
Thus, the answer is \( \boxed{1} \).