Step-by-step Solution:

\[ \begin{aligned} \text{Given circles:} & \quad x^2 + y^2 - 2x - 15 = 0 \\ & \quad x^2 + y^2 - 1 = 0 \end{aligned} \] Step 1: General Equation of Circle Let the equation of the required circle be: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] Since the circle passes through \( (0,1) \), we substitute: \[ 1 + 2f + c = 0 \] \[ \Rightarrow c = -1 - 2f \] Step 2: Condition for Orthogonality For two circles \( x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 \) and \( x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 \) to be orthogonal, the condition is: \[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \] # Applying Orthogonality Condition with the First Circle: Comparing \( x^2 + y^2 - 2x - 15 = 0 \) with \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we get: \[ g_1 = -1, \quad f_1 = 0, \quad c_1 = -15, \quad g_2 = g, \quad f_2 = f, \quad c_2 = c \] \[ 2(-1)g + 2(0)f = -15 + c \] \[ -2g = c - 15 \] # Applying Orthogonality Condition with the Second Circle: Comparing \( x^2 + y^2 - 1 = 0 \) with \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we get: \[ g_1 = 0, \quad f_1 = 0, \quad c_1 = -1 \] \[ 2(0)g + 2(0)f = -1 + c \] \[ 0 = c - 1 \Rightarrow c = 1 \] Step 3: Solving for \( g \) and \( f \) From \( c = 1 \), substituting in \( -2g = c - 15 \): \[ -2g = 1 - 15 = -14 \] \[ g = 7 \] From \( 1 + 2f + c = 0 \): \[ 1 + 2f + 1 = 0 \] \[ 2f = -2 \Rightarrow f = -1 \] Step 4: Finding Centre and Radius The centre of the required circle is \( (-g, -f) = (-7,1) \). The radius is given by: \[ r = \sqrt{g^2 + f^2 - c} \] \[ r = \sqrt{7^2 + (-1)^2 - 1} \] \[ r = \sqrt{49 + 1 - 1} = \sqrt{49} = 7 \] Final Answer The required circle has: \[ \text{Centre: } (-7,1), \quad \text{Radius: } 7 \]