Question

The equation of the circle, the end points of whose diameter are the centre of the circles $x^2+y^2+6x-14y=1$ and $x^2+y^2-4x+10y=2$ is

• A. $x^2+y^2+x-2y-14=0$
• B. $x^2+y^2+x+2y-14=0$
• C. $x^2+y^2+x+2y+14=0$
• D. $x^2+y^2+x-2y=0$

Answer 1

The correct option is A $x^2+y^2+x-2y-14=0$

The general equation of circle is $x^2+y^2+2gx+2fy+c=0$

where,

Center = $(-g,-f)$

Radius = $\sqrt{g^2+f^2-c}$

So,

For $x^2+y^2+6x-14y=1$,

Center = $(-3,7)$

And For $x^2+y^2-4x+10y=2$,

Center = $(2,-5)$

Since centers of given circles are diametric ends of new circle,

For new circle,

Center = $(\frac{-3+2}{2},\frac{7-5}{2})=(\frac{-1}{2},1)$

Radius = $\frac{\sqrt{(-3-2{)}^2+(7+5{)}^2}}{2}=\frac{13}{2}$

Hence, Equation of new circle

$\to (x+\frac{1}{2}{)}^2+(y-1{)}^2=\frac{169}{4}$

$\to x^2+y^2+x-2y-41=0$