# The equation of a circle whose center is at ( 7 , − 3 ) and is tangent to x -axis.

### 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 99E
To determine

## To calculate: The equation of a circle whose center is at (7,−3) and is tangent to x -axis.

Expert Solution

The equation of the circle is (x7)2+(y+3)2=9 .

### Explanation of Solution

Given information:

The center of the circle is at (7,3) and is tangent to x -axis.

Formula used:

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.

Distance d(A,B) between two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by d(A,B)=(x2x1)2+(y2y1)2 .

Calculation:

Consider the provided conditions that center of the circle is at (7,3) and is tangent to x -axis.

Since the circle is tangent to x -axis so it touches the x -axis at a single point and x -coordinate of the center is same as the x -coordinates of point on x -axis.

Also y coordinate will be zero when circle is tangent to x -axis.

So, the circle passes through the point (7,0) .

Now, the distance between the point through which the circle passes and the center of the circle is radius.

Recall that the distance d(A,B) between two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by d(A,B)=(x2x1)2+(y2y1)2 .

Evaluate the distance between (7,0) and (7,3) .

d=(77)2+(30)2=9=3

Therefore, radius of circle is 3.

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.

Compare, (h,k) and (7,3) also radius r is 3.

Here, h=7,k=3 and r=3 .

Substitute the values in standard equation of circle,

(x7)2+(y(3))2=32(x7)2+(y+3)2=9

Thus, the equation of circle is (x7)2+(y+3)2=9 .

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