Question

Find the equation of the circle the end points of whose diameter are the centres of the circles $x^2+y^2+16x-14y=1$ and $x^2+y^2-4x+10y=2$

• A. $x^2+y^2+6x-2y-51=0$
• B. $x^2+y^2+10x-2y-19=0$
• C. $x^2+y^2-6x+2y-51=0$
• D. $x^2+y^2-10x+2y-51=0$

Answer 1

The correct option is A $x^2+y^2+6x-2y-51=0$

Let $C_1$ and $C_2$ be the centres of the circles $x^2+y^2+16x-14y=1$ and $x^2+y^2-4x+10y=2$
$\Rightarrow C_1=(-8,7)$ and $C_2=(2,-5)$
Equation of circle with $C_1$ and $C_2$ as end point s of diametre is
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$
$\Rightarrow (x+8)(x-2)+(y-7)(y+5)=0$
$\Rightarrow x^2+6x-16+y^2-2y-35=0$
$\therefore$ Required circle equation is $x^2+y^2+6x-2y-51=0$
Hence, option A.