Question

If \( x - 4 = 0 \) is the radical axis of two orthogonal circles out of which one is \( x^2 + y^2 = 36 \), then the centre of the other circle is:

• A. \( (8,0) \)
• B. \( (9,0) \)
• C. \( (6,0) \)
• D. \( (12,0) \)

Hint:

For two circles to be orthogonal, they must satisfy the equation \( 2g_1 g_2 + 2f_1 f_2 = r_1^2 + r_2^2 \). The radical axis helps in identifying the second circle's center.

Answer 1

The Correct Option is B

We are given: - One circle: \( x^2 + y^2 = 36 \) with center \( (0, 0) \) and radius \( 6 \). - The radical axis of the two circles is the line \( x - 4 = 0 \). Step 1: Equation of the Second Circle Let the second circle have the general form: \[ (x - h)^2 + y^2 = r^2 \] Since the given radical axis is \( x - 4 = 0 \), by definition of the radical axis: \[ \text{(Equation of first circle)} - \text{(Equation of second circle)} = 0 \] Substituting the known circle equation, \[ x^2 + y^2 - [(x - h)^2 + y^2 - r^2] = 0 \] Expanding, \[ x^2 + y^2 - (x^2 - 2hx + h^2 + y^2 - r^2) = 0 \] Simplifying, \[ x^2 + y^2 - x^2 + 2hx - h^2 - y^2 + r^2 = 0 \] \[ 2hx - h^2 + r^2 = 0 \] Since the radical axis is \( x - 4 = 0 \), the equation must be in the form \( 2hx = h^2 - r^2 + 16 \). Equating the linear term with the radical axis equation: \[ 2h = 1 \quad \Rightarrow \quad h = 9 \] Step 2: Identify the Centre The center of the second circle is \( (9, 0) \). Step 3: Final Answer 

\[Correct Answer: (2) \ (9, 0)\]