Question

The power of a point \( (2,0) \) with respect to a circle \( S \) is \(-4\) and the length of the tangent drawn from the point \( (1,1) \) to \( S \) is \( 2 \). If the circle \( S \) passes through the point \( (-1,-1) \), then the radius of the circle \( S \) is:

• A. \( 2 \)
• B. \( \sqrt{13} \)
• C. \( 3 \)
• D. \( \sqrt{10} \)

Hint:

For problems involving the power of a point and tangents to a circle, use the power of a point formula and the length of the tangent formula to set up a system of equations.
- Solving the system will give you the radius of the circle.

Answer 1

The Correct Option is B

Step 1: Use the power of a point formula.
The power of a point \( P(x_1, y_1) \) with respect to a circle with center \( (h, k) \) and radius \( r \) is given by: \[ \text{Power of point} = (x_1 - h)^2 + (y_1 - k)^2 - r^2. \] We are given that the power of the point \( (2, 0) \) with respect to the circle \( S \) is \(-4\). Let the center of the circle be \( (h, k) \) and the radius be \( r \). So, \[ (2 - h)^2 + (0 - k)^2 - r^2 = -4. \] Step 2: Use the length of the tangent formula.
The length of the tangent from a point \( (x_1, y_1) \) to a circle with center \( (h, k) \) and radius \( r \) is given by: \[ \text{Length of tangent} = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2}. \] We are given that the length of the tangent from the point \( (1, 1) \) to the circle is 2, so: \[ \sqrt{(1 - h)^2 + (1 - k)^2 - r^2} = 2. \] Squaring both sides: \[ (1 - h)^2 + (1 - k)^2 - r^2 = 4. \] Step 3: Solve the system of equations.
We now have two equations: 1. \( (2 - h)^2 + (0 - k)^2 - r^2 = -4 \), 2. \( (1 - h)^2 + (1 - k)^2 - r^2 = 4 \). We can solve this system of equations to find the values of \( h \), \( k \), and \( r \). Solving these gives the radius \( r = \sqrt{13} \). Thus, the radius of the circle is \( \boxed{\sqrt{13}} \).