Ellipse

Concept:When two curves intersect, any curve passing through their intersection points can be expressed as a linear combination of their equations.For the combined curve to be a circle, the coefficients of $x^2$ and $y^2$ must be equal.Explanation:Step 1: Write the given ellipses.$S_1: x^2 + 2y^2 - 6x - 12y + 23 = 0$$S_2: 4x^2 + 2y^2 - 20x - 12y + 35 = 0$Step 2: Form the family $S_1 + \lambda S_2 = 0$.$(x^2 + 2y^2 - 6x - 12y + 23) + \lambda(4x^2 + 2y^2 - 20x - 12y + 35) = 0$This simplifies to:$(1+4\lambda)x^2 + (2+2\lambda)y^2 - (6+20\lambda)x - (12+12\lambda)y + (23+35\lambda) = 0$Step 3: Apply condition for a circle: coefficient of $x^2$ = coefficient of $y^2$.$1 + 4\lambda = 2 + 2\lambda$Solve: $2\lambda = 1 \Rightarrow \lambda = \frac{1}{2}$Step 4: Substitute $\lambda = \frac{1}{2}$ into the combined equation.$3x^2 + 3y^2 - 16x - 18y + \frac{81}{2} = 0$Divide by 3: $x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0$Step 5: Find centre $(a,b)$ from the circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$.$2g = -\frac{16}{3} \Rightarrow g = -\frac{8}{3}$$2f = -6 \Rightarrow f = -3$Centre $= (-g, -f) = \left(\frac{8}{3}, 3\right)$So $a = \frac{8}{3}$, $b = 3$.Step 6: Find radius $r$ using $r^2 = g^2 + f^2 - c$.$r^2 = \left(-\frac{8}{3}\right)^2 + (-3)^2 - \frac{27}{2}$$r^2 = \frac{64}{9} + 9 - \frac{27}{2}$Common denominator 18: $r^2 = \frac{128 + 162 - 243}{18} = \frac{47}{18}$Step 7: Compute $ab + 18r^2$.$ab = \left(\frac{8}{3}\right)(3) = 8$$18r^2 = 18 \times \frac{47}{18} = 47$$ab + 18r^2 = 8 + 47 = 55$Answer:55 (Option C)