Question

The equation of the circle which passes through the origin and cuts orthogonally each of the circles
\[ x^2+y^2-6x+8=0 \] and
\[ x^2+y^2-2x-2y=7 \] is

• A. \(3x^2+3y^2-8x-13y=0\)
• B. \(3x^2+3y^2-8x+29y=0\)
• C. \(3x^2+3y^2+8x+29y=0\)
• D. \(3x^2+3y^2-8x-29y=0\)

Hint:

Circle through origin: \(x^2+y^2+2gx+2fy=0\). Orthogonality with \(x^2+y^2+2g_1x+2f_1y+c_1=0\) gives \(2gg_1+2ff_1=c_1\).

Answer 1

The Correct Option is B

Step 1: General circle through origin.
\[ x^2+y^2+2gx+2fy=0 \]
Step 2: Condition for orthogonality.
Two circles \(S=0\) and \(S_1=0\) are orthogonal if:
\[ 2g_1g_2+2f_1f_2=c_1+c_2 \]
Step 3: Write both given circles in standard form.
Circle 1:
\[ x^2+y^2-6x+8=0 \Rightarrow 2g_1=-6\Rightarrow g_1=-3,\; f_1=0,\; c_1=8 \]
Circle 2:
\[ x^2+y^2-2x-2y-7=0 \Rightarrow 2g_2=-2\Rightarrow g_2=-1,\; 2f_2=-2\Rightarrow f_2=-1,\; c_2=-7 \]
Step 4: Apply orthogonality with unknown circle.
Unknown circle: \(g,f,c=0\).
With circle 1:
\[ 2g(-3)+2f(0)=0+8 \Rightarrow -6g=8 \Rightarrow g=-\frac{4}{3} \]
With circle 2:
\[ 2g(-1)+2f(-1)=0-7 \Rightarrow -2g-2f=-7 \Rightarrow g+f=\frac{7}{2} \]
Substitute \(g=-\frac{4}{3}\):
\[ -\frac{4}{3}+f=\frac{7}{2} \Rightarrow f=\frac{7}{2}+\frac{4}{3} =\frac{21+8}{6}=\frac{29}{6} \]
Step 5: Form equation.
\[ x^2+y^2+2gx+2fy=0 \Rightarrow x^2+y^2-\frac{8}{3}x+\frac{29}{3}y=0 \]
Multiply by 3:
\[ 3x^2+3y^2-8x+29y=0 \]
Final Answer:
\[ \boxed{3x^2+3y^2-8x+29y=0} \]