Circle Geometry, SMB-002 The line \(AT\) is the tangent to the circle at \(A\), and \(BT\) is a secant meeting the circle at \(B\) and \(C\). Given that \(AT = 12\), \(BC = 7\) and \(CT = x\), find the value of \(x\). (2 marks) --- 5 WORK AREA LINES (style=lined) --- Show Answers Only \(x = 9\) Show Worked Solution \(\text{Property: square of tangent = product of secant intercepts}\) \(AT^2\) \(= CT \times BT\) \(12^2\) \(= x(x + 7)\) \(144\) \(= x^2 + 7x\) \(x^2 + 7x-144\) \(= 0\) \((x + 16)(x-9)\) \(= 0\) \(\therefore x = 9,\ (x \gt 0) \)