Question

The centre of the circles $x^2-y^2-6x-8y-7=0$ and $x^2-y^2-4x-10y-3=0$ are the ends of the diameter of the circle:

• A. $x^2-y^2-5x-9y-26=0$
• B. $x^2-y^2-5x-9y+26=0$
• C. $x^2+y^2+5x-y-14=0$
• D. $x^2+y^2+5x+y+14=0$

Answer 1

The correct option is A $x^2-y^2-5x-9y-26=0$

Let $C_1:x^2+y^2-6x-8y-7=0$

$\therefore (x^2-6x)+(y^2-8y)-7=0$

$\therefore (x^2-6x+9)-9+(y^2-8y+16)-16-7=0$

$\therefore {(x-3)}^2+{(y-4)}^2=32$

$\therefore {(x-3)}^2+{(y-4)}^2={\left(\sqrt{32}\right)}^2$

Comparing this equation with the standard equation of circle i.e. ${(x-h)}^2+{(y-k)}^2=r^2$, we get,

$h=3$ or $k=4$ and $r_1=\sqrt{32}$

Thus, coordinates of center of circle $C_1$ are $A(3,4)$ and radius is $\sqrt{32}$

Let $C_2:x^2+y^2-4x-10y-3=0$

$\therefore (x^2-4x)+(y^2-10y)-3=0$

$\therefore (x^2-4x+4)-4+(y^2-10y+25)-25-3=0$

$\therefore {(x-2)}^2+{(y-5)}^2=32$

$\therefore {(x-2)}^2+{(y-5)}^2={\left(\sqrt{32}\right)}^2$

Comparing this equation with standard form of equation of circle i.e. ${(x-h)}^2+{(y-k)}^2=r^2$, we get,

$h=2$, $k=5$ and $r_2=\sqrt{32}$

Thus, coordinates of center of $C_2$ are $B(2,5)$ and radius is $\sqrt{32}$

Now, this line AB is diameter of another circle $C_3$. Mid-point of this line will be center of circle $C_3$

Let coordinates of center of circle $C_3$ are $C(h,k)$

Thus, by mid-point formula, we can write,

$h=\frac{3+2}{2}$

$\therefore h=\frac{5}{2}$

Similarly, $k=\frac{4+5}{2}$

$\therefore k=\frac{9}{2}$

Thus, coordinates of center are $C(\frac{5}{2},\frac{9}{2})$

Now, as shown in figure, CA and CB are radii of circle $C_3$

Thus, by distance formula,

$CA=\sqrt{{(3-\frac{5}{2})}^2+{(4-\frac{9}{2})}^2}$

$\therefore CA=\sqrt{{\left(\frac{1}{2}\right)}^2+{(-\frac{1}{2})}^2}$

$\therefore CA=\sqrt{\left(\frac{1}{4}\right)+\left(\frac{1}{4}\right)}$

$\therefore CA=\sqrt{\frac{1}{2}}$

$\therefore CA=\frac{1}{\sqrt{2}}$

Thus, equation of circle $C_3$ is given by,

${(x-\frac{5}{2})}^2+{(y-\frac{9}{2})}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$

$\therefore x^2-5x+\frac{25}{4}+y^2-9y+\frac{81}{4}=\frac{1}{2}$

$\therefore x^2+y^2-5x-9y+\frac{106}{4}=\frac{1}{2}$

$\therefore x^2+y^2-5x-9y=\frac{1}{2}-\frac{106}{4}$

$\therefore x^2+y^2-5x-9y=-\frac{104}{4}$

$\therefore x^2+y^2-5x-9y=-26$

$\therefore x^2+y^2-5x-9y+26=0$

Thus, answer is option (B)