Question

If a circle \(S\) passes through \((a, b)\) and cuts the circle \(x^2 + y^2 = 4\) orthogonally, then find the locus of the center of \(S\).

• A. \(2ax - 2by + (a^2+b^2+4) = 0\)
• B. \(2ax + 2by - (a^2+b^2+4) = 0\)
• C. \(2ax + 2by + (a^2+b^2+4) = 0\)
• D. \(2ax - 2by - (a^2+b^2+4) = 0\)

Hint:

For a circle \(S\) cutting a circle \(x^2+y^2=R^2\) orthogonally, the power of the center of \(S\) with respect to the circle is \(R^2\).

Answer 1

The Correct Option is B

Step 1: Write circle $S$ in general form.
Let $S$ have centre $(h,k)$ and radius $r$: \[ x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0, \] so $g_1=-h$, $f_1=-k$, $c_1=h^2+k^2-r^2$.
Step 2: Write the fixed circle.
The circle $x^2+y^2=4$ has $g_2=0$, $f_2=0$, $c_2=-4$.
Step 3: Apply the orthogonality condition.
Orthogonal circles satisfy $2g_1g_2+2f_1f_2=c_1+c_2$. The left side is $0$, so \[ 0=(h^2+k^2-r^2)-4\ \Longrightarrow\ r^2=h^2+k^2-4. \]
Step 4: Use that $S$ passes through $(a,b)$.
Then $(a-h)^2+(b-k)^2=r^2$. Substitute $r^2$ from Step 3: \[ (a-h)^2+(b-k)^2=h^2+k^2-4. \]
Step 5: Expand and simplify.
Expanding, $a^2-2ah+b^2-2bk=-4$, which rearranges to $2ah+2bk-(a^2+b^2+4)=0$.
Step 6: Replace centre with $(x,y)$.
The locus of the centre is \[ 2ax+2by-(a^2+b^2+4)=0, \] which is option 2.
\[ \boxed{2ax+2by-(a^2+b^2+4)=0} \]