Question

The equation of the circle which cuts orthogonally each of the three circles given below:
$x^2+y^2-2x+3y-7=0$, $x^2+y^2+5x-5y+9=0$ and $x^2+y^2+7x-9y+29=0$

• 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 $x^2+y^2-16x-18y-4=0$

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

Alternate Solution:
$S_1:x^2+y^2-2x+3y-7=0S_2:x^2+y^2+5x-5y+9=0S_3:x^2+y^2+7x-9y+29=0$
radical axis for $S_1$ and $S_2$
$S_2-S_1=0\Rightarrow 7x-8y+16=0\cdots \left(1\right)$
radical axis for $S_2$ and $S_3$
$S_3-S_2=0\Rightarrow x-2y+10=0\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ radical centre of the given circle will be
$(8,9)$
Hence equation of the required circle will be
$(x-8{)}^2+(y-9{)}^2=(\sqrt{\left(S_1\right)}{)}^2\Rightarrow x^2+y^2-16x-18y-4=0$