The conditions on coefficients a, b and c such that equation x 2 + a x + y 2 + b y + c = 0 represents a circle, a single point and an empty set and evaluate the center and radius of the circle.

Precalculus: Mathematics for Calcu...

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

Precalculus: Mathematics for Calcu...

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

Solutions

Chapter 1.8, Problem 120E
To determine

To calculate: The conditions on coefficients a, b and c such that equation x2+ax+y2+by+c=0 represents a circle, a single point and an empty set and evaluate the center and radius of the circle.

Expert Solution

The equation x2+ax+y2+by=c represents a circle when a2+b24c4>0 , a point when a2+b24c4=0 and an empty when a2+b24c4<0 . The center of the circle is (a2,b2) and radius is a2+b24c4 .

Explanation of Solution

Given information:

The equation x2+ax+y2+by+c=0 .

Formula used:

In order to solve a quadratic equation, completing the square method is used that transforms the equation in the form of square trinomial.

Step1: Divide the equation by coefficient of x2 if it is not equal to 1.

Step 2: Take the square of the half of the coefficient of x and add it to both sides of the equation.

Step 3: Factor the equation.

The standard form of the equation of the circle is (xh)2+(yk)2=r2 , where (h,k) denote the center of the circle and r denote the radius.

Calculation:

Consider the equation x2+ax+y2+by+c=0 .

Recall that in order to solve a quadratic equation, completing the square method is used that transforms the equation in the form of square trinomial.

Step1: Divide the equation by coefficient of x2 if it is not equal to 1.

Step 2: Take the square of the half of the coefficient of x and add it to both sides of the equation.

Step 3: Factor the equation.

Isolate the constant term to right hand side of the equation,

x2+ax+y2+by=c

Next, add a24 and b24 to both the sides of the equation,

x2+ax+y2+by+a24+b24=c+a24+b24

Group the terms,

x2+ax+y2+by+a24+b24=c+a24+b24(x2+ax+a24)+(y2+by+b24)=a2+b24c4

Factor out the trinomial, recall that (a+b)2=a2+2ab+b2 and (ab)2=a22ab+b2 .

Apply it,

x2+ax+y2+by+a24+b24=c+a24+b24(x2+ax+a24)+(y2+by+b24)=a2+b24c4(x+a2)2(y+b2)2=a2+b24c4

Now, the above expression represent a circle when the term on the right hand side is strictly greater than 0 that is a2+b24c4>0 .

Also, the above expression represent a point when the term on the right hand side is equal to 0 that is a2+b24c4=0 that is sum of squares of two numbers is equal to 0.

And the above expression represent an empty set when the term on the right hand side is strictly less than 0 that is a2+b24c4<0 that is sum of squares of two numbers is negative which is not possible.

Recall that the standard form of the equation of the circle is (xh)2+(yk)2=r2 , where (h,k) denote the center of the circle and r denote the radius.

Convert the equation obtained above in standard form,

(x+a2)2(y+b2)2=a2+b24c4(x(a2))2(y(b2))2=(a2+b24c4)2

Compare, (xh)2+(yk)2=r2 and (x(a2))2(y(b2))2=(a2+b24c4)2 .

Here, h=a2,k=b2 and r=a2+b24c4 .

Therefore, center of circle is (a2,b2) and radius is a2+b24c4 .

Thus, the equation x2+ax+y2+by=c represents a circle when a2+b24c4>0 , a point when a2+b24c4=0 and an empty when a2+b24c4<0 . The center of the circle is (a2,b2) and radius is a2+b24c4 .

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