A circle of equation x2+y2+ax+by+c= 0 passes through (0, 6) and touches y=x2 at (2, 4). Find a+c.

Answer: \(a + c = 16\)

1. Passes through (0, 6):

\[36 + 6b + c = 0 \quad\Rightarrow\quad 6b + c = -36 \ \ (1)\]

2. Passes through (2, 4):

\[4 + 16 + 2a + 4b + c = 0\]

\[2a + 4b + c = -20 \ \ (2)\]

3. Touches \(y = x^2\) at (2, 4): slope of the parabola there is \(\frac{dy}{dx} = 2x = 4\). Differentiating the circle,

\[2x + 2yy' + a + by' = 0 \Rightarrow y' = -\frac{2x + a}{2y + b}\]

At (2, 4) this must equal 4:

\[-\frac{4 + a}{8 + b} = 4 \Rightarrow a + 4b = -36 \ \ (3)\]

4. Solve. From (3), \(a = -36 - 4b\). Put in (2):

\[-72 - 8b + 4b + c = -20 \Rightarrow c = 52 + 4b\]

Put in (1): \(6b + 52 + 4b = -36\), so \(b = -8.8\).

Then \(c = 52 - 35.2 = 16.8\) and \(a = -36 + 35.2 = -0.8\).

\[a + c = -0.8 + 16.8 = 16\]