# The equation for the given circle with center ( − 5 , − 1 ) and which also passes through the origin. ### 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, Problem 105RE
To determine

## To calculate:The equation for the given circle with center (−5,−1) and which also passes through the origin.

Expert Solution

The equation for the given circle with center (5,1) and passing through origin is:

x2+y2+10x+2y=0

### Explanation of Solution

Given information:

A circle with center (h,k)=(5,1) and it passes through origin.

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.

But, when an extra condition that is the circle passes through origin is imposed on the standard equation then with the help of distance formula to get D between two points say (x1,y1) and (x2,y2) is given as:

D=(x2x1)2+(y2y1)2

Then we get r2=h2+k2

Putting this in standard equation we get:

The equation for the given circle with center (h,k) and which also passes through the origin is:

(xh)2+(yk)2=h2+k2

Calculation:

Consider the given circle with center (h,k)=(5,1) and passing through origin.

Recall that the equation for the given circle with center (h,k) and which also passes through the origin is:

(xh)2+(yk)2=h2+k2

Therefore, replacing the values we get:

(x(5))2+(y(1))2=(5)2+(1)2(1)

Now using the following formulae:

(a+b)2=a2+2ab+b2(ab)2=a22ab+b2

Then, (1) reduces to:

x2+25+10x+y2+1+2y=25+1x2+y2+10x+2y+26=26x2+y2+10x+2y+2626=0x2+y2+10x+2y=0

Hence, the equation for the given circle with center (5,1) and passing through origin is:

x2+y2+10x+2y=0

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