To write: The standard form of the equation of the circle that passes through the points with the coordinates (1,4),(2,−1),(−3,0) . Then identify the center and radius.

Expert Solution & Answer

Answer to Problem 33E

(x+16)2+(y−76)2=16918r=1326Center(−16,76)

Explanation of Solution

Given information: The coordinates (1,4),(2,−1),(−3,0)

Calculation:

Points (x,y):(1,4),(2,−1),(−3,0)

Plug in the 3 points to create 3 equations which we can use to solve the system to find all of the missing values.

Remember the form:

(x−h)2+(y−k)2=r2Vertex at:(h,k)

Pt:(1,4):(1−h)2+(4−k)2=r2h2+k2−2h−8k+17=r2Pt:(2,−1):(2−h)2+(−1−k)2=r2h2+k2−4h+2k+5=r2Pt:(−3,0):(−3−h)2+(0−k)2=r2h2+k2+6h+9=r2

Set 1^s^t 2 equations equal to each other: because they both equal r2 .

h2+k2−2h−8k+17=h2+k2+6h+98h=−8k+8→h=−k+1

Set the 1^s^t and 3^r^d equations equal: because they also both equal r2 .

h2+k2−2h−8k+17=h2+k2−4h+2k+52h=10k−12→h=5k−6

Set these 2 results equal because they both equal r²

−k+1=5k−6→k=76→h=−16

You know the center (h,k)=(−16,76)

Plug these values for h and k into any of the equations to find r .

(−16)2+(76)2+6(−16)+9=r2r2=16918

Standard form:

(x+16)2+(y−76)2=16918r=1326Center(−16,76)

Chapter 10.2, Problem 32E

Chapter 10.2, Problem 34E

Chapter 10 Solutions

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

Ch. 10.1 - Prob. 1CFU Ch. 10.1 - Prob. 2CFU Ch. 10.1 - Prob. 3CFU Ch. 10.1 - Prob. 4CFU Ch. 10.1 - Prob. 5CFU Ch. 10.1 - Prob. 6CFU Ch. 10.1 - Prob. 7CFU Ch. 10.1 - Prob. 8CFU Ch. 10.1 - Prob. 9CFU Ch. 10.1 - Prob. 10CFU

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