Question

I'm not sure where I'm going wrong please help

Answer 1

A)

so as you already know, this is just a matter of distance, thus we use the distance formula from the center (-1 , 5) to a point with an x = 0 or (0 , y) with a distance of 6.

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{y})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ 6= √((~~ 0- (-1) ~~)^2 + (~~ y- 5 ~~)^2)\implies 6^2=(0+1)^2+(y^2-10y+25) \\\\\\ 36=y^2-10y+26\implies 0=y^2-10y-10 \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{1}y^2\stackrel{\stackrel{b}{\downarrow }}{-10}y\stackrel{\stackrel{c}{\downarrow }}{-10} \qquad \qquad y= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}`

`y= \cfrac{ - (-10) \pm \sqrt { (-10)^2 -4(1)(-10)}}{2(1)} \implies y = \cfrac{ 10 \pm \sqrt { 100 +40}}{ 2 } \\\\\\ y= \cfrac{ 10 \pm \sqrt { 140 }}{ 2 }\implies y= \cfrac{ 10 \pm 2\sqrt { 35 }}{ 2 }\implies y=5\pm √(35)\implies \begin{cases} (0,5-√(35))\\\\ (0,5+√(35)) \end{cases}`

B)

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{x}~,~\stackrel{y_2}{2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ 6= √((~~ x- (-1) ~~)^2 + (~~ 2- 5 ~~)^2)\implies 6^2=(x+1)^2 + (-3)^2 \\\\\\ 36=x^2+2x+1+9\implies 0=x^2+2x-26 \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+2}x\stackrel{\stackrel{c}{\downarrow }}{-26} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}`

`x= \cfrac{ - (2) \pm \sqrt { (2)^2 -4(1)(-26)}}{2(1)} \implies x = \cfrac{ -2 \pm \sqrt { 4 +104}}{ 2 } \\\\\\ x= \cfrac{ -2 \pm \sqrt { 108 }}{ 2 }\implies x= \cfrac{ -2 \pm 6\sqrt { 3 }}{ 2 }\implies x=-1\pm 3√(3)~\hfill~ \begin{cases} (-1+3√(3),2)\\\\ (-1-3√(3),2) \end{cases}`