Chapter 9, Problem 4CT

To determine

To find equation for the set of all points equidistant from point $\left(2,0\right)$ and line $y=-4$ .

Answer to Problem 4CT

The equation for the set of all points equidistant from point $\left(2,0\right)$ and line $y=-4$ is $x^2+y^2-4x-12=0$ .

Explanation of Solution

Given:

Point $=$$\left(2,0\right)$

Line $y=-4$

Formula Used:

Circle equation is given as ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ ; where $\left(h,k\right)$ is the Center and $r$ is the radius.

Calculation:

Given:

Point $=$$\left(2,0\right)$

Line $y=-4$

Thus, we have:

Center $=$$\left(2,0\right)$

Radius $=-4$

Using the above formula and putting the above values in the equation of the Circle, we have:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

  ${\left(x-2\right)}^2+{\left(y-0\right)}^2={\left(-4\right)}^2$

Solving the above equation, we have:

  $\begin{array}{l}{\left(x-2\right)}^2+{\left(y-0\right)}^2={\left(-4\right)}^2\\ x^2+4-4x+y^2=16\\ x^2+y^2-4x+4-16=0\\ x^2+y^2-4x-12=0\end{array}$

Equation of the Circle is $x^2+y^2-4x-12=0$ .

Thus, the equation for the set of all points equidistant from point $\left(2,0\right)$ and line $y=-4$ is $x^2+y^2-4x-12=0$ .

Conclusion:

The equation for the set of all points equidistant from point $\left(2,0\right)$ and line $y=-4$ is $x^2+y^2-4x-12=0$ .