The point \(A\) has coordinates \((5,16)\) and the point \(B\) has coordinates \((-4,4)\). The variable point \(P\) has coordinates \((x,y)\) and moves on a path such that \(AP = 2BP\). Show that the Cartesian equation of the path of \(P\) is \[(x + 7)^2 + y^2 = 100.\]
The point \(C\) has coordinates \((a,0)\) and the point \(D\) has coordinates \((b,0)\). The variable point \(Q\) moves on a path such that \[QC = k \times QD,\] where \(k > 1\). Given that the path of \(Q\) is the same as the path of \(P\), show that \[\frac{a + 7}{b + 7} = \frac{a^2 + 51}{b^2 + 51}.\] Show further that \((a + 7)(b + 7) = 100\), in the case \(a \neq b\).
Geometry of Equations
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