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# Intersection(s) of graph and line

Q: The graph of $$y=\frac 4x$$ and a line through $$[0,4]$$ may intersect.
What slope of the line results in 2,1 or 0 intersections?
Where does the line touch the graph in just 1 point?

#### Graph:

\begin{align} \frac 4x &=mx+4 &|\times x \\ 4 &=mx^2+4x &|-4 \\ mx^2 + 4x - 4 &=0 &|\div m \\ x^2 + \frac 4m x - \frac 4m & =0 \\ \left(x + \frac 2m\right)^2 - \left(\frac 2m\right)^2 - \frac 4m & =0 \\ \left(x + \frac 2m\right)^2 & = \frac{4}{m^2} + \frac 4m & =0 \\ x & = - \frac 2m \pm \sqrt{\frac{4}{m^2} + \frac 4m} \\ x & = - \frac 2m \pm \sqrt{\frac 4m \left(\frac 1m + 1\right)} \\ \end{align} \\
$$\frac 4m \left(\frac 1m + 1 \right) = 0 \text{ results in 1 real solution exactly when } m = -1\\ \frac 4m \left(\frac 1m + 1\right) >0 \text{ results in 2 real solution exactly when } m < -1\\ \frac 4m \left(\frac 1m + 1\right) <0 \text{ results in 0 real solution exactly when } m > -1\\ \text{With }m=-1 \\ x=-\frac{2}{-1} \pm \sqrt{\frac{4}{-1}\left(\frac{1}{-1}+1\right)} \\ x=2 \pm \sqrt{(-4) \times \left(-1+1\right)} \\ x=2 \pm 0 \\ x=2 \\ y=\frac 4x =\frac 42 = 2\\ A=[2,2] \\$$

# Tangents of line on graph

Q: The graph of $$y=x^2-2x+2$$ and a line through $$[1,0]$$ may intersect. What slope(s) of the line results in exactly 1 intersection?
Where do the lines touch the graph in just 1 point?

#### Graph:

\begin{align} y & = x^2-2x+2, \text{ and } \\ y & = mx+k, \text{ with } y(+1)=0 \implies 0=m(+1) + k \implies k=-m \implies \\ y & = mx-m \\ \end{align} \\
Graph and line intersect when:
\begin{align} mx-m &= x^2-2x+2 \\ x^2-mx-2x+m+2 &= 0 \\ x^2-(m+2)x+(m+2) &= 0 \\ \left(x-\frac{m+2}{2}\right)^2 - \left(\frac{m+2}{2}\right)^2 + (m+2) &= 0 \\ \left(x-\frac{m+2}{2}\right)^2 &= + \frac{(m+2)^2}{4} -m-2 \\ x-\frac{m+2}{2} &= \pm \sqrt{\frac{m^2 +4m+4}{4} -m-2} \\ x &= \frac{m+2}{2} \pm \sqrt{\frac{m^2 +4m+4}{4} +\frac{-4m-8}{4}}\\ x &= 1+\frac{m}{2} \pm \frac 12 \sqrt{m^2 +4m+4 -4m -8}\\ x &= 1+\frac{m}{2} \pm \frac 12 \sqrt{m^2 -4}\\ A=[2,2] \\ B=[0,2] \\ \end{align} \\
This term has a single solution $$x$$ if
\begin{align} m^2-4 &= 0\\ m^2 &= 4\\ m&=\pm\sqrt{4} \\ m &=\pm 2 \\ m_1 &=2 \\ m_2 &=-2 \\ \end{align} \\ y_1=f_1(x)=+2x-2 \implies x=1+1=2 \implies y(2)=2 \\ y_1=f_1(x)=-2x+2 \implies x=1+(-1)=0 \implies y(0)=2 \\

# Intersecting parabolas

Q: For which $$m$$ do the graphs of $$y=x^2-2x+2$$ and $$y=-x^2+m$$ may just touch in one single point?

#### Graph:

\begin{align} x^2-2x+2 &= -x^2+m &&| +x^2 -m \\ 2x^2-2x-m+2 &= 0 &&| \div 2 \\ x^2-x-\frac{m-2}{2} &&= 0 \\ \left(x-\frac 12\right)^2-\left(\frac 12\right)^2 -\frac{m-2}{2} &= 0 \\ \left(x-\frac 12\right)^2 &=\left(\frac 12\right)^2 +\frac{m-2}{2} \\ x &=+\frac 12 \pm \sqrt{\left(\frac 12\right)^2 +\frac{m-2}{2}} \\ \end{align}
There is exactly one solution if:
\begin{align} \sqrt{\left(\frac 12\right)^2 +\frac{m-2}{2}}&=0 &&\lvert \text{ square} \\ \left(\frac 12\right)^2 +\frac{m-2}{2}&=0 &&\lvert-\left(\frac 12\right)^2 \\ \frac{m-2}{2}&=-\frac 14 &&| \times 2 \\ m-2&=-\frac 12 &&| +2 \\ m&=1\frac 12 \\ \end{align}

# Gradient on parabola point (algebraic solution)

Q: What is the function of the line touching the graphs of $$y=x^2$$ at $$P=[1,1]$$?

#### Graph:

\begin{align} y&=mx+b\\ \text{ Any line going through } [1,1] \text{ must fulfil: }\\ 1&=m\cdot 1+b &&| -m \\ b&=-m+1 && (*) \\ \text{ Therefore any line going through } [1,1] \text{ is: }\\ y&=mx-m+1 \\ \text{ Find } m \text{ for which is there only 1 solution to: } \\ x^2&=mx-m+1 &&|-mx \\ x^2-mx &=-m+1 &&| \text{ C.T.S.} \\ \left(x-\frac m2 \right)^2 - \left(\frac m2 \right)^2 &=-m+1 &&| +\left(\frac m2 \right)^2 \\ \left(x-\frac m2 \right)^2 &=-m+1 +\left(\frac m2 \right)^2 &&| \sqrt{} \\ x&=+\frac m2 \pm \sqrt{-m+1 +\frac {m^2}{4}} \\ \text{ There's exactly one solution if the root expression is } 0 \\ \frac{m^2}{4}-m+1&=0 &&| \times 4 \\ m^2-4m+4&=0 \\ (m-2)^2&=0 \\ m&=2 &&| \text{ use } (*)\\ b&=-m+1=-1 \\ y&=2x-1 \\ \end{align} \\
A: The tangent function is $$y=2x-1$$.

# Points on Circle

The circle $$C$$ passes through the points $$P$$, $$Q$$ and $$R$$ with coordinates $$(-2,-2)$$, $$(2,-4)$$ and $$(7,1)$$ respectively.

Q (a): Find an equation of the perpendicular bisector of the points $$P$$ and $$Q$$.

Q (b): Find the coordinates of the centre of $$C$$.

Q (c): Find an equation of $$C$$.

#### Answer a: perpendicular bisector of PQ:

For PQ with $$P=(-2,-2)$$ and $$Q=(2,-4)$$
the perpendicular bisector is
$$y=mx+b$$ with m being the negative inverse slope of the line PQ:
$$m=-\frac{1}{\frac{-4-(-2))}{2-(-2)}}=-\frac{1}{\frac{-2}{4}}=-\frac{1}{-0.5}=2$$
$$y=2x+b$$
with the midpoint of PQ as $$(0,-3)$$ on that line, there is
$$-3=2(0)+b \\ b=-3$$
and
$$\mathbf{y=2x-3}$$

#### Answer b: circle center (C)

Similarly, the halfway point between $$Q=(2,-4), R=(7,1)$$ is
$$(\frac{2+7}{2},\frac{-4+1}{2})=(4.5,-1.5) \\$$ and the gradient is
$$-\frac{1}{\frac{7-2}{1-(-4)}}=-1 \\$$
Therefore the line equation is
$$y=-x+b$$
with $$-1.5=-4.5+b$$, therefore
$$b=3$$ , therefore the perpendicular bisector of QR is:
$$\mathbf{y=-x+3}$$
Those 2 lines intersect (at $$C$$ ) where:
$$-x+3=2x-3 \\ 6=3x \\ x=2 \\ y=-x+3=-2+3=1$$
Therefore the centre $$C$$ of the circle is
\mathbf{C=(2,1)} \).

The squared radius $$r^2$$ of the circle with $$C=(2,1)$$ and $$R=(7,1)$$:
$$r^2=(2-7)^2 + (1-1)^2=25$$
Therefore the circle equation is:
$$\mathbf{(x-2)^2+(y-1)^2=25}$$

# Parallel Tangent

The circle $$C$$ has equation $$x^2+y^2-4x-6=0$$ and the line $$l$$ has equation $$y=3x-6$$.

Q (a): Show that $$l$$ passes through the centre of $$C$$.

Q (b): Find an equation for each tangent to $$C$$ that is parellel to $$l$$.

#### Answer a: $$l$$ passing through centre of circle:

\begin{align} x^2+y^2-4x-6&=0 \\ (x-2)^2-4+y^2-6&=0 \\ (x-2)^2+y^2-10&=0 \\ (x-2)^2+y^2&=10 \\ \end{align}

The circle has its centre at $$(2,0)$$.

\begin{align} 3x-6&=0 \\ 3\cdot2-6&=0 \\ 0&=0 \checkmark \\ \end{align}

The line goes through $$(2,0)$$.

#### Answer b: equations of tangents parallel to $$l$$

A parallel line to $$l$$ is $$y=3x+t$$.

If $$y=3x+t$$ intersects the circle, $$(x,y)$$ are identical for the circle and line equation at those points, i.e. $$y$$ can be substituted.

\begin{align} x^2+(3x+t)^2-4x-6&=0 \\ x^2+9x^2+6xt+t^2-4x-6&=0 \\ 10x^2+(6t-4)x+t^2-6&=0 \\ x^2+\frac{(3t-2)x}{5}+\frac{t^2-6}{10}&=0 \\ \left(x-\frac{3t-2}{10}\right)^2-\left(\frac{3t-2}{10}\right)^2+\frac{t^2-6}{10}&=0 \\ x&=+\frac{3t-2}{10}\pm\sqrt{\left(\frac{3t-2}{10}\right)^2-\frac{t^2-6}{10}} && (*) \\ \end{align}

In order for the line $$3x+t$$ to touch the circle in only one point, $$x$$ has to be a singular solutions in $$(*)$$, i.e.

\begin{align} \left(\frac{3t-2}{10}\right)^2-\frac{t^2-6}{10}&=0 \\ \frac{9t^2-12t+4}{100}-\frac{t^2-6}{10}&=0 \\ 9t^2-12t+4-10t^2+60&=0 \\ -t^2-12t+64&=0 \\ t^2+12t-64&=0 \\ (t+6)^2-36-64&=0 \\ (t+6)^2-100&=0 \\ t&=-6\pm10 \\ y_1&=3x+4 \\ y_2&=3x-16 \\ \end{align}