#1**+4 **

I think that this question is fairly difficult. I have added a few extra lines and points to the diagram where I saw fit:

Here are my initial observations in regard to this current diagram:

- The x-axis, which is blue, is tangent to both circles.
- The green line is also tangent to both circles.
- The green and blue lines intersect at one common point on the x-axis.

These initial observations are enough to get me started with this problem. I constructed a parallel line to the x-axis, \(\overleftrightarrow{AK}\). \(\angle BKA\) is, therefore, a right angle. Point K is located at (8,1) because it acquires the y-coordinate of Point A and the x-coordinate of Point B. \(\triangle BKA\) is a right triangle, and it is possible to find \(\tan(m\angle KAB)\).

\(\tan m\angle KAB=\frac{9-1}{8-2}=\frac{8}{6}=\frac{4}{3}\). Notice how 4/3 is also the slope of \(\overleftrightarrow{AB}\). The actual measure of the angle is not really important.

Because \(\overleftrightarrow{AK}||\overleftrightarrow{JH}\), \(\angle HJB\cong\angle KAB\). By extension, \(m\angle HJB=m\angle KAB\). This also means that \(\tan m\angle HJB=\tan m\angle KAB=\frac{4}{3}\).

Because two tangent lines of a circle intersect at an external point, J, \(\angle IJB\cong\angle HJB\). Using the same reasoning as before, \(\tan m\angle IJB=\tan m\angle HJB=\frac{4}{3}\). \(m\angle IJH=m\angle IJB+m\angle HJB=m\angle HJB+m\angle HJB=2m\angle HJB\)

Both tangent lines meet at an angle of \(2m\angle HJB\). Just like finding \(\tan m\angle KAB\) yielded the slope, \(\tan 2m\angle HJB\) will yield the slope of the desired line, \(\overleftrightarrow{IJ}\).

Luckily, we can utilize the Double-Angles Identities, specifically the \(\tan 2m\angle HJB=\frac{2\tan m\angle HJB}{1-\tan^2 m\angle HJB}\).

\(\tan 2m\angle HJB=\frac{2\tan m\angle HJB}{1-\tan^2 m\angle HJB};\\ \tan m\angle HJB=\frac{4}{3}\) | This is the Double-Angle Identity for the tangent function. Replace all instances of \(\tan m\angle HJB\) with 4/3. |

\(\tan 2m\angle HJB=\frac{2*\frac{4}{3}}{1-\left(\frac{4}{3}\right)^2}\\ \tan 2m\angle HJB=\frac{\frac{8}{3}}{\frac{-7}{9}}\\ \tan 2m\angle HJB=\frac{-24}{7} \) | Simplify. The result is the slope of the line \(\overleftrightarrow{IJ}\) . |

\(y=\frac{-24}{7}x+b\) | This is the equation of the line \(\overleftrightarrow{IJ}\). |

The equation of the circle with the center A is \((x-2)^2+(y-1)^2=1\). Since \(y=\frac{-24}{7}x+b\), substitute that into the equation for the circle:

\((x-2)^2+(y-1)^2=1;\\ y=\frac{-24}{7}x+b \) | These are the equations we are grappling with. |

\((x-2)^2+(\frac{-24}{7}x+[b-1])^2=1\) | It is time to expand this monstrosity. |

\(x^2-4x+4+\left(\frac{-24}{7}x\right)^2+2*\frac{-24}{7}x*(b-1)+(b-1)^2=1\) | |

\(x^2-\frac{28}{7}x+4+\frac{576}{49}x^2-\frac{48b-48}{7}x+b^2-2b+4=0\) | |

\(\textcolor{red}{\frac{625}{49}}x^2\textcolor{blue}{-\left(\frac{48b-20}{7}\right)}x+\textcolor{green}{(b^2-2b+4)}=0\\ \textcolor{red}{a}x^2+\textcolor{blue}{b}x+\textcolor{green}{c}=0\) | Notice the parallelism between a quadratic and the current equation. Tangent lines always intersect a circle at only one point, so let's set the discriminant equal to zero to ensure only one solution is possible. |

\(\Delta=b^2-4ac;\\ \Delta=\left(-\frac{48b-20}{7}\right)^2-4*\frac{625}{49}*(b^2-2b+4)\\ \Delta=\frac{-196b^2+3080b-9600}{49}\) | Set the discriminant equal to zero and solve. |

\(\frac{-196b^2+3080b-9600}{49}=0\Rightarrow-196b^2+3080b-9600=0\) | Use the quadratic formula here. |

\(b_1=\frac{30}{7}\text{ or }b_2=\frac{80}{7}\) | Reject b_2 because that is on the other side of the circle. |

TheXSquaredFactor
16-nov-2018

#1**+2 **

The maximum or minimum of any quadratic is always at the input value of \(\frac{-b}{2a}\) where the standard form of a quadratic is \(ax^2+bx+c=0\) .

\(a=-4, b=16;\\ x=\frac{-b}{2a}\) | Substitute in the known values for a and b . |

\(x=\frac{-16}{2*-4}\) | Solve for x by simplifying completely. |

\(x=2\) | This is the optimal number of items sold, in thousands, in order to make maximum profit. Let's substitute that into the original function to determine the profit with that number of items sold. |

\(x=2;\\ p(2)=-4*2^2+16*2-7\) | Yet again, just simplify from here. |

\(p(2)=-4*4+32-7\) | |

\(p(2)=-16+25\) | |

\(p(2)=9\) | This is the maximum profit that this business, in thousands of dollars, makes. |

TheXSquaredFactor
14-oct-2018

#1**+1 **

Hi, nellycrane!

I think you will be better off if I give you an easier problem like \(x^2-4 \) . You are probably well aware that this is a difference of squares. \((x+2)(x-2)\) would be the corresponding factorization.

Now, let's say that you want to factor \(x^2-30\) . This is not possible if you restrict yourself to the rational number set. However, it can be factored as \((x+\sqrt{30})(x-\sqrt{30})\).

In general, \(a^2-b^2=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) . You can apply this knowledge to the problems at hand.

\(x^2+50\\ a=x^2,b=-50;\\ (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\\ (\sqrt{x^2}+\sqrt{-50})(\sqrt{x^2}-\sqrt{-50}) \)

You can simplify this to get the factorization amongst the complex numbers. Good luck!

TheXSquaredFactor
29-sep-2018

#1**+2 **

This problem is easier than it looks at first glance. If you utilize clever algebraic manipulation, this problem becomes simpler.

\(x-y=16\) | \(xy=23\) |

\((x-y)^2=16^2\) | |

\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\) | \(\boxed{2}\hspace{1mm}2xy=46\) |

Notice what I have done. I have manipulated the information I know about these real numbers, *x *and *y* , and I am manipulating it in a way that is much more convenient for this particular problem. The only thing left to do is add the equations together.

\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\\ \boxed{2}\hspace{5mm}+2xy\hspace{10mm}=46\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \hspace{18mm}x^2+y^2=302\)

TheXSquaredFactor
29-sep-2018