Chapter 10.1, Problem 17PSB

(a)

To determine

To find: the radius of the circle from the given points on circle.

Answer to Problem 17PSB

Radius $=13$ .

Explanation of Solution

Given:

  

Concept used:

A tangent to the circle is a line interesting the circle at exactly one point tangentially to the circumference of the circle an important result come out is radius from the centre of the circle to the point of tangency is perpendicular to the tangent line.

Calculation:

According to the given:

  $x$ axis is a tangent.

  $\Rightarrow PN\perp x$ axis.

If a line is tangent to a circle, it is perpendicular to the radius drawn to the point of tangency or the point of contact.

Hence, Radius $=13$ .

(b)

To determine

To find: the length or distance of the $PQ$ .

Answer to Problem 17PSB

The distance is $5$ unit.

Explanation of Solution

Given:

  

Concept used:

Distance formula:

  $PQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

According to the distance formula:

  $PQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

By putting the value:

  $PQ=\sqrt{{\left(15-19\right)}^2+{\left(13-16\right)}^2}=\sqrt{16+9}=5.$

Hence, the distance is $5$ unit.

(c)

To determine

To find: the distance or length of the segment $AB$ .

Answer to Problem 17PSB

The length of the chord is $24$ .

Explanation of Solution

Given:

  

Concept used:

If a radius of circle bisects a chord that is not a diameter then it is perpendicular to that chord.

Pythagoras theorem can be used for right angle triangle:

  $H^2=P^2+B^2$ .

Calculation:

  $\begin{array}{l}AQ=QB.\\ \Rightarrow PQ\perp AB.\end{array}$

If a radius of circle bisects a chord that is not a diameter then it is perpendicular to that chord.

  $\begin{array}{l}\Rightarrow BQ=\sqrt{P{B}^2-P{Q}^2}=\sqrt{{13}^2-5^2}=\sqrt{144}=12.\\ \Rightarrow AB=24.\end{array}$

Hence, the length of the chord is $24$ .