What is the shortest distance from the origin to the circle defined by \(x^2-24x +y^2+10y +160=0\)?
The equation can also be written as \(\left(x-12\right)^{2\ }+\ \left(y+5\right)^2=9\), so we know that the radius of the circle is 3, and the origin is at (12,-5).
+6248 I think the easiest way to do this is to parameterize the circle via a single variable and solve the minimization problem.
\(\text{let }p \text{ be a point on the circle given by }(x-12)^2 + (y+5)^2 = 9 \\ \text{then p can be described as }\\ p=(3 \cos(\theta)+12,~3\sin(\theta)-5)\)
\(\text{The squared distance of this point to the origin is }\\ d^2 = (3 \cos(\theta)+12)^2 + (3\sin(\theta)-5)^2 =\\ 9 \sin ^2(\theta )-30 \sin (\theta )+9 \cos ^2(\theta )+72 \cos (\theta )+169 = \\ -30 \sin (\theta )+72 \cos (\theta )+178 = \\ 78 \cos(\theta+\arctan(5/12))+178\)
\(d^2 \text{will clearly be minimized when }\\ \theta + \arctan(5/12) = \pi\\ \theta = \pi - \arctan(5/12)\\ \text{at this angle }d^2 = -78 + 178 = 100\\ d=10\)
