Question

NO LINKS!! Please help me with this problem. Find all points on the x-axis that are a distance 5 from P(-8,4)​

Answer 1

`(x,y)=\left(\; \boxed{-11,0}\; \right)\; \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{-5,0}\; \right)\; \textsf{(larger $x$-value)}`

Explanation:

`\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}}`

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:

`\implies (x+8)^2+(y-4)^2=25`

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:

`\implies (x+8)^2+(0-4)^2=25`

`\implies (x+8)^2+(-4)^2=25`

`\implies (x+8)^2+16=25`

`\implies (x+8)^2+16-16=25-16`

`\implies (x+8)^2=9`

`\implies √((x+8)^2)=√(9)`

`\implies x+8=\pm3`

`\implies x+8-8=\pm3-8`

`\implies x=-8\pm3`

`\implies x=-11, x=-5`

Therefore:

`(x,y)=\left(\; \boxed{-11,0}\; \right)\; \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{-5,0}\; \right)\; \textsf{(larger $x$-value)}`