Answer:
[tex](x,y)=\left(\; \boxed{-11,0}\; \right)\; \textsf{(smaller $x$-value)}[/tex]
[tex](x,y)=\left(\; \boxed{-5,0}\; \right)\; \textsf{(larger $x$-value)}[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{4 cm}\underline{Equation of a circle}\\\\$(x-a)^2+(y-b)^2=r^2$\\\\where:\\ \phantom{ww}$\bullet$ $(a, b)$ is the center. \\ \phantom{ww}$\bullet$ $r$ is the radius.\\\end{minipage}}[/tex]
The points that are a distance of 5 units from P(-8, 4) will be all the points on the circumference of a circle with center (-8, 4) and radius 5.
Substitute the center and radius into the formula to create an equation of the circle:
[tex]\implies (x+8)^2+(y-4)^2=25[/tex]
The y-value of any point on the x-axis is zero.
Therefore, to find the points on the x-axis that are 5 units from point P, substitute y = 0 into the equation of the circle and solve for x:
[tex]\implies (x+8)^2+(0-4)^2=25[/tex]
[tex]\implies (x+8)^2+(-4)^2=25[/tex]
[tex]\implies (x+8)^2+16=25[/tex]
[tex]\implies (x+8)^2+16-16=25-16[/tex]
[tex]\implies (x+8)^2=9[/tex]
[tex]\implies \sqrt{(x+8)^2}=\sqrt{9}[/tex]
[tex]\implies x+8=\pm3[/tex]
[tex]\implies x+8-8=\pm3-8[/tex]
[tex]\implies x=-8\pm3[/tex]
[tex]\implies x=-11, x=-5[/tex]
Therefore:
[tex](x,y)=\left(\; \boxed{-11,0}\; \right)\; \textsf{(smaller $x$-value)}[/tex]
[tex](x,y)=\left(\; \boxed{-5,0}\; \right)\; \textsf{(larger $x$-value)}[/tex]
