Chapter 12.7, Problem 42SPR

(a)

To determine

To find the location of 4^th tower that is equidistant from the given towers

Answer to Problem 42SPR

The location of 4^th tower that is equidistant from the given towers is $\left(3,-4\right)$

Explanation of Solution

Given information:

Three radio towers are modeled by the points $A\left(-3,4\right),B\left(9,4\right)$ and $C\left(-3,-12\right)$ .

It can be represented in a Cartesian plane as follows:

Formula used:

Circumcenter: The circumcenter of a triangle is the point in the plane which is equidistant from the three vertices of thetriangle.

To find the location of 4^th tower that is equidistant from the given three towers means that we have to find the circumcenter of the triangle formed by the 3 vertices.

The circumcenter will be the intersection of all perpendicular bisectorsof sides of the triangle.

To find perpendicular vectors:

Midpoint of the points $A\left(-3,4\right)\text{and C}\left(9,4\right)$ = $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$=\left(\frac{-3+9}{2},\frac{4+4}{2}\right)$

$=\left(3,4\right)$

$$

Midpoint of the points $A\left(-3,4\right)\text{and B}\left(-3,-12\right)$

$=\left(\frac{-3+-3}{2},\frac{4-12}{2}\right)$

$=\left(-3,-4\right)$

Draw perpendicular lines through midpoints as shown in the diagram:

$$

From the diagram, we get, the intersection of perpendicular bisector of the lines AB and AC, circumcenter, is the point $\left(3,-4\right)$

That is, the location of 4^th tower that is equidistant from the given towers at $A\left(-3,4\right),B\left(9,4\right)$ and $C\left(-3,-12\right)$ is $\left(3,-4\right)$

(b)

To determine

To find the equation for the circle which all the given 3 vertices lies.

Answer to Problem 42SPR

Equation of circle is ${\left(x-3\right)}^2+{\left(y+4\right)}^2=100$

Explanation of Solution

Given information:

Three radio towers are modeled by the points $A\left(-3,4\right),B\left(9,4\right)$ and $C\left(-3,-12\right)$ .

Formula Used:

• Standard form of equation of circle:

The standard form of the equation of circle with center at $\left(h,k\right)$ and radius r is

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

• Distance between the points $\left(x_1,y_1\right)\text{and }\left(x_2,y_2\right)$is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

To find the equation of circle with center at $\left(3,-4\right)$ ,and which lies all three vertices $A\left(-3,4\right),B\left(9,4\right)$ and $C\left(-3,-12\right)$ ,we need to find the radius.

To find the radius:

Since the vertices A, B and C lies on the circle, distance between the center $\left(3,-4\right)$ and any one of the point A,B, and C will be radius of the circle.

Radius, r = distance between the points $\left(3,-4\right)$ and $\left(-3,4\right)$

= $\sqrt{{\left((-3)-3\right)}^2+{\left(4-(-4)\right)}^2}$

= $\sqrt{{\left(-6\right)}^2+{\left(8\right)}^2}$

  $=\sqrt{36+64}$

  $=\sqrt{100}$

  $=10$

Equation of circle with center $\left(3,-4\right)$ and radius 10 is

${\left(x-3\right)}^2+{\left(y-(-4)\right)}^2={10}^2$

${\left(x-3\right)}^2+{\left(y+4\right)}^2=100$