Question

If \( x - 2y - 6 = 0 \) is a normal to the circle \( x^2 + y^2 + 2gx + 2fy - 8 = 0 \) and the line \( y = 2 \) touches this circle, then the radius of the circle can be:

• A. \( \sqrt{32} \)
• B. 6
• C. 4
• D. \( \sqrt{18} \) \vspace{0.5cm}

Hint:

When solving problems involving tangency and normal lines, always use the geometric properties of the circle and line to establish relationships between the radius, center, and the equations of the lines.

Answer 1

The Correct Option is C

We are given that the line \( x - 2y - 6 = 0 \) is normal to the circle. This means that the radius of the circle at the point of contact with this line is perpendicular to it. Also, the line \( y = 2 \) touches the circle, which means the distance from the center of the circle to this line equals the radius of the circle. Step 1: Find the center of the circle The equation of the circle is \( x^2 + y^2 + 2gx + 2fy - 8 = 0 \). To find the center, we rewrite the equation in the standard form \( (x + g)^2 + (y + f)^2 = r^2 \). The center of the circle is at the point \( (-g, -f) \), and the radius of the circle is \( r \). Step 2: Use the normal line equation The line \( x - 2y - 6 = 0 \) is normal to the circle, which means that the line passes through the center of the circle. The center of the circle \( (-g, -f) \) must satisfy the equation of this line. Substituting \( (-g, -f) \) into the equation of the normal line: \[ -g - 2(-f) - 6 = 0 \] \[ -g + 2f - 6 = 0 \] \[ g = 2f - 6 \quad \text{(Equation 1)} \] Step 3: Use the tangency condition The line \( y = 2 \) is tangent to the circle. The distance from the center of the circle \( (-g, -f) \) to the line \( y = 2 \) is equal to the radius of the circle. The distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is given by the formula: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] For the line \( y = 2 \), we rewrite it as \( 0x + 1y - 2 = 0 \). The distance from the center \( (-g, -f) \) to the line is: \[ d = \frac{|0(-g) + 1(-f) - 2|}{\sqrt{0^2 + 1^2}} = \frac{| -f - 2 |}{1} = |f + 2| \] Since this distance is equal to the radius \( r \), we have: \[ |f + 2| = r \] Step 4: Solve the system Now, using Equation 1 and the fact that \( g = 2f - 6 \), substitute \( g \) into the equation for the radius. By solving the system of equations, we find that the radius of the circle \( r = 4 \). Thus, the radius of the circle is \( \boxed{4} \). \vspace{0.5cm}