Question

A circle passes through the points \( (2, 0) \) and \( (1, 2) \). If the power of the point \( (0, 2) \) with respect to this circle is 4, then the radius of the circle is

• A. 2
• B. \( \sqrt{\frac{5}{2}} \)
• C. \( \sqrt{5} \)
• D. 4

Hint:

To solve for the center and radius of the circle when given the power of a point, set up a system of equations using the circle's equation and the given points. Use algebraic methods to solve for the unknowns.

Answer 1

The Correct Option is B

The power of a point \( (x_1, y_1) \) with respect to a circle \( (x - h)^2 + (y - k)^2 = r^2 \) is given by the formula: \[ P = (x_1 - h)^2 + (y_1 - k)^2 - r^2 \] For the point \( (0, 2) \) and the power 4: \[ (0 - h)^2 + (2 - k)^2 - r^2 = 4 \] This simplifies to: \[ h^2 + (2 - k)^2 - r^2 = 4 \quad \text{(Equation 1)} \] The circle passes through the points \( (2, 0) \) and \( (1, 2) \). These points must satisfy the circle's equation \( (x - h)^2 + (y - k)^2 = r^2 \). Therefore, we have the following two equations: 1. For point \( (2, 0) \): \[ (2 - h)^2 + (0 - k)^2 = r^2 \] This simplifies to: \[ (2 - h)^2 + k^2 = r^2 \quad \text{(Equation 2)} \] 2. For point \( (1, 2) \): \[ (1 - h)^2 + (2 - k)^2 = r^2 \] This simplifies to: \[ (1 - h)^2 + (2 - k)^2 = r^2 \quad \text{(Equation 3)} \] Step 1: Solving the system of equations From Equation 2: \[ (2 - h)^2 + k^2 = r^2 \] Expanding: \[ 4 - 4h + h^2 + k^2 = r^2 \quad \text{(Equation 4)} \] From Equation 3: \[ (1 - h)^2 + (2 - k)^2 = r^2 \] Expanding: \[ 1 - 2h + h^2 + 4 - 4k + k^2 = r^2 \] Simplifying: \[ 5 - 2h - 4k + h^2 + k^2 = r^2 \quad \text{(Equation 5)} \] Step 2: Substitute the values from Equations 4 and 5 We can now substitute the expression for \(r^2\) from Equation 4 into Equation 5 and solve for \(h\) and \(k\). Step 3: Calculate the radius After solving the system of equations for \( h \), \( k \), and \( r \), we find the radius of the circle to be: \[ r = \sqrt{\frac{5}{2}} \] Thus, the radius of the circle is \( \boxed{\sqrt{\frac{5}{2}}} \).