Question

The equation of the circle passing through the points of intersection of the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-10x-12y+40=0$ and whose radius is $4$ is ?

• A. $x^2+y^2-2x-15=0$
• B. $x^2+y^2-2y-15=0$
• C. $x^2+y^2-2x+2y-15=0$
• D. $x^2+y^2+2x+2y-15=0$

Answer 1

The correct option is B $x^2+y^2-2y-15=0$

The Equation of the circle passing through the point of intersection of the circle is given by

$\begin{array}{c}{x}^2+y^2-2x-4y-4+\lambda \left(x^2+y^2-10x-12y+40\right)=0\\ \left(1+\lambda \right)x^2+\left(1+\lambda \right)y^2+\left(-2-10\lambda \right)x+\left(-4-12\lambda \right)y-4+40\lambda =0\\ x^2+y^2-\frac{\left(1+10\lambda \right)x}{1+\lambda }-\frac{\left(4+12\lambda \right)y}{1+\lambda }+\frac{\left(-44+40\lambda \right)}{1+\lambda }\\ Oncomparingbothsideswith\\ x^2+y^2+2gx+2fy+c=0\\ weget\\ g=-\frac{\left(1+5\lambda \right)}{1+\lambda },f=-\frac{\left(2+6\lambda \right)}{1+\lambda },c=\frac{-4+40\lambda }{1+\lambda }\\ radius=\sqrt{g^2+f^2-c^2}\\ =4\\ {\left(\frac{1+5\lambda }{1+\lambda }\right)}^2+{\left(\frac{2+6\lambda }{1+\lambda }\right)}^2-\left(\frac{40\lambda -4}{1+\lambda }\right)=16\\ 1+25{\lambda }^2+10\lambda +4+36{\lambda }^2+24\lambda -\left(40\lambda +40{\lambda }^2-4-4\lambda \right)=16{\left(1+\lambda \right)}^2\\ 5{\lambda }^2-34\lambda -7=0\\ 5{\lambda }^2+\lambda -35\lambda -7=0\\ \lambda \left(5\lambda +1\right)-7\left(5\lambda +1\right)=0\\ \left(5\lambda +1\right)\left(\lambda -7\right)=0\\ \lambda =-\frac{1}{5},7\\ Putting\lambda =-\frac{1}{5}intheequation\\ x^2+y^2-2y-15=0\end{array}$

Hence, the correct option is $B$