Question

If $P(\frac{7}{5}, \frac{6}{5})$ is the inverse point of $A(1,2)$ with respect to a circle with centre $C(2,0)$, then the radius of that circle is

• A. 9
• B. 3
• C. $\sqrt{3}$
• D. 1

Hint:

The concept of inverse points is a geometric transformation. The two key properties are collinearity with the center (C, A, P are on a line) and the metric relation $CA \cdot CP = R^2$. You can use this to find the radius or the location of one of the points.

Answer 1

The Correct Option is C

Step 1: State the property of inverse points.
If P is the inverse point of A with respect to a circle centered at C with radius R, then C, A, and P are collinear, and the product of the distances from the center to the points equals the radius squared. \[ CA \cdot CP = R^2. \]

Step 2: Calculate the distance CA.
The points are $C(2,0)$ and $A(1,2)$. \[ CA = \sqrt{(1-2)^2 + (2-0)^2} = \sqrt{(-1)^2 + 2^2} = \sqrt{1+4} = \sqrt{5}. \]

Step 3: Calculate the distance CP.
The points are $C(2,0)$ and $P(7/5, 6/5)$. \[ CP = \sqrt{\left(\frac{7}{5}-2\right)^2 + \left(\frac{6}{5}-0\right)^2} = \sqrt{\left(-\frac{3}{5}\right)^2 + \left(\frac{6}{5}\right)^2}. \] \[ CP = \sqrt{\frac{9}{25} + \frac{36}{25}} = \sqrt{\frac{45}{25}} = \frac{\sqrt{9 \times 5}}{5} = \frac{3\sqrt{5}}{5}. \]

Step 4: Calculate the radius R.
Using the property $R^2 = CA \cdot CP$: \[ R^2 = (\sqrt{5}) \cdot \left(\frac{3\sqrt{5}}{5}\right) = \frac{3 \times 5}{5} = 3. \] Therefore, the radius is: \[ \boxed{R = \sqrt{3}}. \]