Question

The equation(s) of the circle passing through the points of intersection of the circles $x^2+y^2-2x-4y-4=0,$ $x^2+y^2-10x-12y+40=0$ and having radius $4$ units is/are

• A. $2x^2+2y^2-18x-22y+69=0$
• B. $x^2+y^2+28x-22y+19=0$
• C. $x^2+y^2-2y-15=0$
• D. $x^2+y^2+2x-15=0$

Answer 1

Answer: (A), (C)

$x^2+y^2-2y-15=0$

Equation of any circle through the points of intersection of given circles is
$(x^2+y^2-2x-4y-4)+\lambda (x^2+y^2-10x-12y+40)=0$
$\Rightarrow x^2(1+\lambda )+y^2(1+\lambda )-2x(1+5\lambda )-2y(2+6\lambda )-4+40\lambda =0$ $\Rightarrow x^2+y^2-2x\frac{(1+5\lambda )}{(1+\lambda )}-2y\frac{(2+6\lambda )}{(1+\lambda )}+\frac{(40\lambda -4)}{(1+\lambda )}=0\dots \dots \left(1\right)$

Radius of this circle

$\sqrt{{\left(\frac{1+5\lambda }{1+\lambda }\right)}^2+{\left(\frac{2+6\lambda }{1+\lambda }\right)}^2-\left(\frac{40\lambda -4}{1+\lambda }\right)}=4$

$\Rightarrow \frac{(1+5\lambda {)}^2+(2+6\lambda {)}^2-(40\lambda -4)(1+\lambda )}{(1+\lambda {)}^2}=16$

$\Rightarrow 5{\lambda }^2-34\lambda -7=0$

$\Rightarrow (\lambda -7)(5\lambda +1)=0$

$\therefore \lambda =7\text{or}\lambda =-\frac{1}{5}$

Substituting the values of $\lambda$ in $\left(1\right)$, the required circles are

$2x^2+2y^2-18x-22y+69=0$ and
$x^2+y^2-2y-15=0$