Question

A circle C cuts the circles \(x^{2}+y^{2}-4x+6y+4=0\) and \(x^{2}+y^{2}+6x-4y+9=0\) orthogonally. If origin lies on this circle C, then the radius of the circle C is

• A. \(\frac{\sqrt{85}}{2}\)
• B. \(2\sqrt{3}\)
• C. 4
• D. 5

Hint:

For a circle passing through the origin, the constant term \(c\) in the general equation is always zero, which greatly simplifies the orthogonality equations.

Answer 1

The Correct Option is A

Step 1: Set up the unknown circle.
Let the circle $C$ be $x^2+y^2+2gx+2fy+c=0$. We must find its radius given that it cuts two circles orthogonally and passes through the origin.
Step 2: Use the origin condition.
Since $(0,0)$ lies on $C$, substituting gives $c=0$. That immediately simplifies everything.
Step 3: Recall the orthogonality rule.
Two circles cut orthogonally when $2g_1g_2+2f_1f_2=c_1+c_2$. For circle $C$ against each given circle, this becomes one linear equation in $g$ and $f$.
Step 4: Apply it to both given circles.
Circle 1 has $g_1=-2,\,f_1=3,\,c_1=4$, so $2g(-2)+2f(3)=c+4=4$, giving $-2g+3f=2$.
Circle 2 has $g_2=3,\,f_2=-2,\,c_2=9$, so $2g(3)+2f(-2)=c+9=9$, giving $6g-4f=9$.
Step 5: Solve the two equations.
Multiply $-2g+3f=2$ by $3$ to get $-6g+9f=6$, and add to $6g-4f=9$: \[ 5f=15\implies f=3. \] Then $-2g+9=2$ gives $g=\dfrac{7}{2}$.
Step 6: Find the radius.
The radius is $\sqrt{g^2+f^2-c}=\sqrt{\left(\tfrac72\right)^2+3^2-0}=\sqrt{\tfrac{49}{4}+9}=\sqrt{\tfrac{85}{4}}=\dfrac{\sqrt{85}}{2}$.
\[ \boxed{\dfrac{\sqrt{85}}{2}} \]