BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1, Problem 106RE
To determine

To calculate:The equation for the given circle which has points P(2,3) and Q(1,8) . Also, its center is the midpoint of the given segment PQ

Expert Solution

Answer to Problem 106RE

The equation for the circle thathas points P(2,3) and Q(1,8) , center is the midpoint of the given segment PQ is x2+y2x11y+22=0

Explanation of Solution

Given information:

A circle withpoints P(2,3) and Q(1,8) lying inside it and whose center is given by midpoint of PQ

Formula used:

For a given circle with center (h,k) and radius r the equation is given as:

  (xh)2+(yk)2=r2

This is referred to as the Standard form for the equation of a given circle.

The Distance denoted by say, D between any two given points say (x1,y1) and (x2,y2) is given as:

  D=(x2x1)2+(y2y1)2

The midpoint (x,y) for two points (x1,y1) and (x2,y2) is given as:

  (x,y)=(x1+x22,y1+y22)

The relation between d , diameter of circle and r , radius of circle is:

  r=d2

Calculation:

Consider the given circle with points P(2,3) and Q(1,8) lying inside it.

Then in order to apply the midpoint formula for the segment PQ the following should be known:

  x1=2x2=1y1=3y2=8

Recall the midpoint (x,y) for two points (x1,y1) and (x2,y2) is given as:

  (x,y)=(x1+x22,y1+y22)

Therefore,

  (x,y)=(2+(1)2,3+82)=(12,112)

Hence, center of the circle is (x,y)=(12,112)

Now as midpoint of PQ forms the center of the circle, therefore length of PQ gives the diameter for the circle.

And the relation between d , diameter of circle and r , radius of circle is:

  r=d2

Therefore, distance from center (x,y)=(12,112) to P(2,3) gives the radius for the circle.

Recall the distance D between two points (x1,y1) and (x2,y2) is given as:

  D=(x2x1)2+(y2y1)2

As we have:

  x1=12x2=2y1=112y2=3

Therefore,

  D=(212)2+(3112)2=(32)2+(52)2=94+254=344=342

Hence, radius r=342

Also we have center (h,k)=(12,112)

Recall that for a given circle with center (h,k) and radius r the equation is given as:

  (xh)2+(yk)2=r2

Therefore, replacing the values we get:

  (x12)2+(y112)2=(342)2 (1)

Now using the following formulae:

  (a+b)2=a2+2ab+b2(ab)2=a22ab+b2

Then, (1) reduces to:

  (x12)2+(y112)2=(342)2x2+14x+y2+121411y=344x2+y2x11y+(14+1214344)=0x2+y2x11y+(884)=0x2+y2x11y+22=0

Hence, the equation for the circle that has points P(2,3) and Q(1,8) , center is the midpoint of the given segment PQ is x2+y2x11y+22=0

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!