x²+y²-8x-6y=0 & x²+y²+32x+24y=0  বৃত্তদ্বয়ের ছেদ বিন্দুগামী ও বৃত্তদ্বয়ের কেন্দ্র সমূহের সংযোগকারী  রেখার ওপর লম্ব রেখার সমীকরণ হলো -

Explanation:

Another Explanation (5): Let's find the equation of the line passing through the intersection of the two circles and perpendicular to the line joining the centers of the circles. The given equations of the circles are: \[x^2 + y^2 - 8x - 6y = 0 \quad \cdots (1)\] \[x^2 + y^2 + 32x + 24y = 0 \quad \cdots (2)\] Subtracting equation (1) from (2), we get the equation of the common chord: \[(x^2 + y^2 + 32x + 24y) - (x^2 + y^2 - 8x - 6y) = 0\] \[32x + 24y + 8x + 6y = 0\] \[40x + 30y = 0\] \[4x + 3y = 0 \quad \cdots (3)\] The equation of the line passing through the intersection points of the two circles is \(4x + 3y = 0\). 😊 Now, let's find the centers of the circles. For equation (1), the center is \((4, 3)\). 🤓 For equation (2), the center is \((-16, -12)\). 🤩 The slope of the line joining the centers is: \[m = \frac{-12 - 3}{-16 - 4} = \frac{-15}{-20} = \frac{3}{4}\] The slope of the line perpendicular to the line joining the centers is \(-\frac{4}{3}\). 😎 The equation of the common chord is \(4x + 3y = 0\), which has a slope of \(-\frac{4}{3}\). 😮 Since the line passing through the intersection points is \(4x + 3y = 0\), and this line is perpendicular to the line joining the centers, the required equation is \(4x + 3y = 0\). 🥰 Therefore, the equation of the line is \(4x + 3y = 0\). 🎉