2026 ZHSS P1 Q13

2026 ZHSS P1 Q13

Secondary 4
9 marks

A circle has a diameter \(AB\) where \(A\) is the point \((-2,2)\). The point \(P(2,10)\) also lies on the circle. The tangent to the circle at \(B\) has equation \(3y+4x=k\) where \(k\) is a constant.

  1. Find the equation of the perpendicular bisector of \(AP\).[2]
  2. Find the equation of the circle.[5]
  3. Hence, find the value of \(k\).[2]

Solution:

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Answer:(a) \(y=-\dfrac12x+6\) (b) \((x-2)^2+(y-5)^2=25\) (c) \(k=48\)

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