Question

The equation of the circle which orthogonally intersects the circles $x^2+y^2-2x+3y-7=0$, $x^2+y^2+5x-5y+9=0$ and $x^2+y^2+7x-9y+29=0$, is

• A. $x^2+y^2-16x-8y-12=0$
• B. $x^2+y^2-16x-18y-4=0$
• C. $x^2+y^2+16x-18y+4=0$
• D. $x^2+y^2+16x+18y+12=0$

Answer 1

The correct option is B $x^2+y^2-16x-18y-4=0$

$S_1:x^2+y^2-2x+3y-7=0S_2:x^2+y^2+5x-5y+9=0S_3:x^2+y^2+7x-9y+29=0$

Radical axis of $S_1$ and $S_2$ is
$S_2-S_1=0\Rightarrow 7x-8y+16=0\cdots \left(1\right)$

Radical axis of $S_2$ and $S_3$ is
$S_3-S_2=0\Rightarrow x-2y+10=0\cdots \left(2\right)$
Radical center is point of intersection of equation $\left(1\right)$ and $\left(2\right)$, so
$C=(8,9)$

Hence, the equation of the required circle is
$(x-8{)}^2+(y-9{)}^2=(\sqrt{\left(S_1\right)}{)}^2\Rightarrow (x-8{)}^2+(y-9{)}^2=(64+81-16+27-7)\Rightarrow x^2+y^2-16x-18y-4=0$


Alternate Solution:
Let the required circle be
$x^2+y^2+2gx+2fy+c=0\cdots \left(1\right)$

Since, it is orthogonal to three given circles respectively, therefore
$2g\times (-1)+2f\times \frac{3}{2}=c-7\Rightarrow -2g+3f=c-7\dots \dots \left(2\right)2g\times \frac{5}{2}+2f\times (-\frac{5}{2})=c+9\Rightarrow 5g-5f=c+9\dots \dots \left(3\right)2g\times \frac{7}{2}+2f\times (-\frac{9}{2})=c+29\Rightarrow 7g-9f=c+29\dots \dots \left(4\right)$
Solving equations $\left(2\right),\left(3\right)$ and $\left(4\right)$ we get
$g=-8,f=-9,c=-4$
Substituting the values of $g,f,c$ in $\left(1\right)$ then required circle is
$x^2+y^2-16x-18y-4=0$