JEE Main 2012 — Circle Question with Solution

Point $(1,2)$ lies on the circle $x^2+y^2+2x+2y$ $-11=0$, because coordinates of point $(1,2)$ satisfy the equation $$x^2+y^2+2x+2y-11=0$$ Now, $x^2+y^2-4x-6y-21=0$...(i) $$\begin{array}{rl}& x^2+y^2+2x+2y-11=0\\ & 3x+4y+5=0\end{array}$$ From (i) and (iii), $$\begin{array}{r}{x}^2+{\left(-\frac{3x+5}{4}\right)}^2-4x-6\left(-\frac{3x+5}{4}\right)-21=0\\ \Rightarrow 16x^2+9x^2+30x+25-64x\\ +72x+120-336=0\end{array}$$ $$\begin{array}{rl}& \Rightarrow 25x^2+38x-191=0\\ & \text{ From (ii) and (iii), }\\ & x^2+{\left(-\frac{3x+5}{4}\right)}^2+2x+2\left(-\frac{3x+5}{4}\right)-11=0\\ & \Rightarrow 16x^2+9x^2+30x+25+32x\end{array}$$ Thus we get the same equation from (ii) and (iii) as we get from equation (i) and (iii). Hence the point of intersections of (ii) and (iii) will be same as the point of intersections of(i) and (iii). Therefore the circle (ii) passing through the point of intersection of circle(i) and point $(1,2)$ also as shown in the figure.
Hence equation(ii) i.e. $x^2+y^2+2x+2y-11=0$ is the equation of required circle.