Question

If \( Q \) is the inverse point of the point \( P(2, 3) \) with respect to the circle \( x^2 + y^2 - 2x - 2y + 1 = 0 \), then the circle with \( PQ \) as diameter is

• A. \( 3x^2 + 3y^2 - 14x - 16y + 37 = 0 \)
• B. \( x^2 + y^2 - 4x - 6y + 13 = 0 \)
• C. \( 5x^2 + 5y^2 - 16x - 22y + 33 = 0 \)
• D. \( 2x^2 + 2y^2 - 3x - 3y - 11 = 0 \)

Hint:

Find the center and radius of the given circle. Use the property that the center, the point, and its inverse point are collinear and the product of the distances from the center is the square of the radius. Find the coordinates of the inverse point. Then, find the equation of the circle with the given point and its inverse as the endpoints of the diameter.

Answer 1

The Correct Option is C

The equation of the circle is \( S: x^2 + y^2 - 2x - 2y + 1 = 0 \).
Center \( C(1, 1) \), radius \( r = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1 + 1 - 1} = 1 \).
Let \( Q(x_1, y_1) \) be the inverse point of \( P(2, 3) \) with respect to the circle.
The points \( C, Q, P \) are collinear and \( CP \cdot CQ = r^2 \).
Vector \( \vec{CP} = (2 - 1, 3 - 1) = (1, 2) \).
Vector \( \vec{CQ} = (x_1 - 1, y_1 - 1) \).
Since \( C, Q, P \) are collinear, \( \vec{CQ} = \lambda \vec{CP} \) for some scalar \( \lambda \).
\( (x_1 - 1, y_1 - 1) = \lambda (1, 2) = (\lambda, 2\lambda) \) \( x_1 = 1 + \lambda, y_1 = 1 + 2\lambda \).
\( CP = \sqrt{1^2 + 2^2} = \sqrt{5} \).
\( CQ = \sqrt{\lambda^2 + (2\lambda)^2} = \sqrt{5\lambda^2} = |\lambda|\sqrt{5} \).
\( CP \cdot CQ = r^2 \implies \sqrt{5} \cdot |\lambda|\sqrt{5} = 1^2 \implies 5|\lambda| = 1 \implies |\lambda| = \frac{1}{5} \).
Since \( Q \) lies on the segment \( CP \) extended, \( \lambda \) has the same sign.
Vector \( \vec{CP} \) is from \( C \) to \( P \).
\( Q \) is further from \( C \) than \( P \).
So \( \lambda \) should be such that \( CQ>CP \).
The inverse point lies on the line \( CP \) such that \( CQ = r^2 / CP = 1 / \sqrt{5} \).
\( \vec{CQ} = \frac{1}{5} \vec{CP} = (\frac{1}{5}, \frac{2}{5}) \).
\( x_1 - 1 = \frac{1}{5} \implies x_1 = \frac{6}{5} \).
\( y_1 - 1 = \frac{2}{5} \implies y_1 = \frac{7}{5} \).
So, \( Q = (\frac{6}{5}, \frac{7}{5}) \).
The circle with \( PQ \) as diameter has center as the midpoint of \( PQ \) and radius \( \frac{1}{2} PQ \).
Midpoint \( = \left( \frac{2 + 6/5}{2}, \frac{3 + 7/5}{2} \right) = \left( \frac{16/5}{2}, \frac{22/5}{2} \right) = \left( \frac{8}{5}, \frac{11}{5} \right) \).
Radius \( = \frac{1}{2} \sqrt{(2 - 6/5)^2 + (3 - 7/5)^2} = \frac{1}{2} \sqrt{(\frac{4}{5})^2 + (\frac{8}{5})^2} = \frac{1}{2} \sqrt{\frac{16 + 64}{25}} = \frac{1}{2} \frac{\sqrt{80}}{5} = \frac{1}{2} \frac{4\sqrt{5}}{5} = \frac{2\sqrt{5}}{5} \).
Equation: \( (x - \frac{8}{5})^2 + (y - \frac{11}{5})^2 = (\frac{2\sqrt{5}}{5})^2 = \frac{20}{25} = \frac{4}{5} \).
\( x^2 - \frac{16}{5}x + \frac{64}{25} + y^2 - \frac{22}{5}y + \frac{121}{25} = \frac{4}{5} = \frac{20}{25} \).
\( x^2 + y^2 - \frac{16}{5}x - \frac{22}{5}y + \frac{185 - 20}{25} = 0 \) \( x^2 + y^2 - \frac{16}{5}x - \frac{22}{5}y + \frac{165}{25} = 0 \) \( 5x^2 + 5y^2 - 16x - 22y + 33 = 0 \).