COMEDK202510 May 2025Morning ShiftMathematicsCircleActual
Equation of a circle whose area is 154 sq units and having 2 x-3 y+12=0 and x+4 y-5=0 as diameters is
Options
- Ax^2+y^2+6 x-4 y-36=0
- Bx^2-y^2+6 x-4 y-36=0
- Cx^2+y^2-6 x+4 y-36=0
- Dx^2+y^2+6 x-4 y+36=0
Correct answer
A. x^2+y^2+6 x-4 y-36=0
Step-by-step solution
The area of the circle is given as A = r^2 = 154 . Using = 22 7 , we have 22 7 r^2 = 154 , which implies r^2 = 154 7 22 = 7 7 = 49 . Thus, r = 7 . The center of the circle is the intersection of the two diameters 2x - 3y + 12 = 0 and x + 4y - 5 = 0 . From the second equation, x = 5 - 4y . Substituting this into the first equation: 2(5 - 4y) - 3y + 12 = 0 10 - 8y - 3y + 12 = 0 22 - 11y = 0 y = 2 . Substituting y = 2 back into x = 5 - 4y : x = 5 - 4(2) = 5 - 8 = -3 . The center of the circle is (-3, 2) . The equation