Question

The equation of a circle which is coaxal with the circles $2x^2+2y^2-2x+6y-3=0$ and $x^2+y^2+4x+2y+1=0$, being given that the center of the circle to be determined lies on the radical axis of these circles, is

• A. $2(x^2+y^2)+6x+10y-1=0$
• B. $4(x^2+y^2)+6x+10y-1=0$
• C. $2(x^2+y^2)-6x-10y-1=0$
• D. $4(x^2+y^2)-6x-10y-1=0$

Answer 1

The correct option is B $4(x^2+y^2)+6x+10y-1=0$

The given circles are

$S_1:x^2+y^2-x+3y-\frac{3}{2}=0$

and, $S_2:x^2+y^2+4x+2y+1=0$

$\therefore$ The radical axis of the given circles is $S_1-S_2\equiv -5x+y-\frac{5}{2}=0$

or, $10x-2y+5=0$ ...(1)

$\therefore$ Required circle will have the equation of the form

$x^2+y^2+4x+2y+1+\lambda (10x-2y+5)=0$

or, $x^2+y^2+2(2+5\lambda )x+(1-\lambda )y+(1+5\lambda )=0$ ...(2)

Its centre $\equiv (-2-5\lambda ,\lambda -1).$

Since the centre lies on (1),

$\therefore 10(-2-5\lambda )-2(\lambda -1)+5=0$

or, $-52\lambda -13=0$ or $\lambda =-\frac{1}{4}.$

Putting this value of $\lambda$ in (2), we get

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

or, $x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{1}{4}=0.$

or, $4(x^2+y^2)+6x+10y-1=0.$