Question

Consider two circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-10x-12y+40=0$ Then the centre of the circle passing through the intersection of these two circles with radius $4$ is

• A. $(0,1)$
• B. $(\frac{9}{2},\frac{11}{2})$
• C. $(1,0)$
• D. $(\frac{11}{2},\frac{9}{2})$

Answer 1

The correct option is B $(\frac{9}{2},\frac{11}{2})$

$(x^2+y^2-2x-4y-4)+\lambda (x^2+y^2-10x-12y+40)=0$

$⟹x^2+y^2-2x(\frac{1+5\lambda }{1+\lambda })-2y(\frac{2+6\lambda }{1+\lambda })+\frac{40\lambda -4}{1+\lambda }=0$

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

$⟹\lambda =\frac{-1}{5},7$

$⟹(0,1),(\frac{9}{2},\frac{11}{2})$