See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(for Q. No. 48 and 49) x + 2y + a = 0, 12x – 6y – 41 = 0 and radical axis of the given circles will be concurrent  a = 2 Now for required circle S = 0 x2 + y2 – 4 + (x + 2y + 2) = 0 (Equation of family of circles passing through the intersection points of x + 2y + a=0 and x2+y2=4) Now, common chord with C2 is ( + 4)x + 2( + 1)y + 2 – 5 = 0 Which is same as 12x – 6y – 41 = 0  8 5 