The equation of the circle given is:
\(16(x^2 + y^2) + 32x - 8y - 50 = 0\)
First, simplify the circle's equation:
Next, convert this into the standard circle form \((x - h)^2 + (y - k)^2 = r^2\) by completing the square.
Substitute these adjustments back into the equation:
\((x + 1)^2 - 1 + (y - \frac{1}{4})^2 - \frac{1}{16} - \frac{25}{8} = 0\)
Combine and simplify the constant terms:
\((x + 1)^2 + (y - \frac{1}{4})^2 = \frac{100}{64} = \frac{25}{16}\)
Now, the standard form of the circle is:
\((x + 1)^2 + (y - \frac{1}{4})^2 = \left(\frac{5}{4}\right)^2\)
This shows a circle centered at \((-1, \frac{1}{4})\) with radius \(\frac{5}{4}\).
To find the point on the line \(y = 2x + 11\) closest to the circle's center, use the distance formula. The distance \(d\) from a point \((x_1, y_1)\) to a line \(ax + by + c = 0\) is:
\(d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\)
For the line \(y = 2x + 11\) or (-1, \frac{1}{4}) into the distance formula: \(d = \frac{|2(-1) - 1/4 + 11|}{\sqrt{4 + 1}}\)
To find the specific point on the line, considering the perpendicular slope to \(y = 2x + 11\), an optimal approach involves finding such that the line from \((-1, \frac{1}{4})\) to the point is perpendicular.
The perpendicular slope is \(\frac{-1}{2}\).
Solving simultaneous equations:
Thus, the point on the line \(y = 2x + 11\) nearest to the circle is:
\(\left(-\frac{9}{2}, 2\right)\)