Chapter 13.7, Problem 20PSB

(a)

To determine

To find: The value of PT.

Answer to Problem 20PSB

The value of PT is $5$ .

Explanation of Solution

Given:

The equation of the circle is ${\left(x-4\right)}^2+{\left(y+2\right)}^2=50$ .

The centre of the circle is P and T is the point on the chord line AB such that line PT is perpendicular to line AB.

The coordinates of $A=\left(11,-1\right)$ and $B=\left(5,-9\right)$ .

Calculation:

Consider the midpoint of AB is,

  $\begin{array}{c}T=\left(\frac{11+5}{2},\frac{-1-9}{2}\right)\\ =\left(8,-5\right)\end{array}$

Then, the distance between the point $\left(8,-5\right)$ and P is,

  $\begin{array}{c}PT=\sqrt{\left(8-4\right)+{\left(-5-\left(-2\right)\right)}^2}\\ =5\end{array}$

(b)

To determine

To find: The value of the angle $m\angle TPA$ .

Answer to Problem 20PSB

The value of $m\angle TPA=45°$ .

Explanation of Solution

Consider the given triangle has angle T as the right angle, then the triangles is isosceles and the value of the angle,

  $m\angle TPA=45°$