Question

If the inverse point of the point \(P(3,3)\) with respect to the circle \(x^2 + y^2 - 4x + 4y + 4 = 0\) is \(Q(a,b)\), then \(a + 5b =\)

• A. 4
• B. 0
• C. -4
• D. 1

Hint:

Inversions with respect to degenerate circles (points) typically reflect the original point through the center if the radius is not zero, so check calculations for errors if results seem unexpected.

Answer 1

The Correct Option is C

Let the equation of the circle be \(S: x^2 + y^2 - 4x + 4y + 4 = 0\).
The center of the circle is \(C(2, -2)\).
The radius of the circle is \(r = \sqrt{(-2)^2 + 2^2 - 4} = \sqrt{4 + 4 - 4} = \sqrt{4} = 2\).
Let \(P(3, 3)\) and \(Q(a, b)\) be inverse points with respect to the circle.
Then \(C, P, Q\) are collinear and \(CP \cdot CQ = r^2\).
The slope of \(CP\) is \(\frac{3 - (-2)}{3 - 2} = \frac{5}{1} = 5\).
The equation of the line \(CP\) is \(y + 2 = 5(x - 2)\), i.e., \(y = 5x - 12\).
Since \(Q(a, b)\) lies on this line, \(b = 5a - 12\).
Now, \(CP = \sqrt{(3-2)^2 + (3-(-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{26}\).
\(CQ = \sqrt{(a-2)^2 + (b+2)^2}\).
We have \(CP \cdot CQ = r^2\), so \(\sqrt{26} \cdot \sqrt{(a-2)^2 + (b+2)^2} = 2^2 = 4\).
\((a-2)^2 + (b+2)^2 = \frac{16}{26} = \frac{8}{13}\). Substituting \(b = 5a - 12\):
\((a-2)^2 + (5a - 12 + 2)^2 = \frac{8}{13}\)
\((a-2)^2 + (5a - 10)^2 = \frac{8}{13}\)
\(a^2 - 4a + 4 + 25a^2 - 100a + 100 = \frac{8}{13}\)
\(26a^2 - 104a + 104 = \frac{8}{13}\)
\(13(26a^2 - 104a + 104) = 8\)
\(338a^2 - 1352a + 1352 = 8\)
\(338a^2 - 1352a + 1344 = 0\)
\(169a^2 - 676a + 672 = 0\)
Since \(CP \cdot CQ = r^2\), we have \(\sqrt{26} \cdot \sqrt{(a-2)^2 + (b+2)^2} = 4\).
Also, \(CP = \sqrt{26}\).
We have \(CQ = \frac{4}{\sqrt{26}}\).
Let \(\vec{CP} = \begin{pmatrix} 1
5 \end{pmatrix}\). Then \(\vec{CQ} = \frac{CQ}{CP} \vec{CP} = \frac{4/\sqrt{26}}{\sqrt{26}} \begin{pmatrix} 1
5 \end{pmatrix} = \frac{4}{26} \begin{pmatrix} 1
5 \end{pmatrix} = \frac{2}{13} \begin{pmatrix} 1
5 \end{pmatrix}\).
So, \(\vec{CQ} = \begin{pmatrix} a-2
b+2 \end{pmatrix} = \begin{pmatrix} 2/13
10/13 \end{pmatrix}\).
\(a - 2 = \frac{2}{13} \implies a = 2 + \frac{2}{13} = \frac{28}{13}\).
\(b + 2 = \frac{10}{13} \implies b = \frac{10}{13} - 2 = \frac{10 - 26}{13} = -\frac{16}{13}\).
Then \(a + 5b = \frac{28}{13} + 5\left(-\frac{16}{13}\right) = \frac{28 - 80}{13} = \frac{-52}{13} = -4\). Final Answer: The final answer is $\boxed{(3)}$