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 1.8, Problem 100E
To determine

To calculate: The equation of a circle whose radius is 5 and is tangent to both x -axis and y -axis and lies in first quadrant.

Expert Solution

Answer to Problem 100E

The equation of the circle is (x5)2+(y5)2=25 .

Explanation of Solution

Given information:

The radius of the circle is 5 and is tangent to both x -axis and y -axis and lies in first quadrant.

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 radius of the circle is 5 and is tangent to both x -axis and y -axis and lies in first quadrant.

Since the circle is tangent to x -axis so it touches the x -axis at a single point and y -coordinate is 0.

As the radius of circle is 5 units so point on x -axis through which the circle passes is (5,0) .

Since the circle is tangent to y -axis so it touches the y -axis at a single point and x -coordinate is 0.

As the radius of circle is 5 units so point on y -axis through which the circle passes is (0,5) .

So, the center of the circle is (5,5) .

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 (5,5) also radius r is 5.

Here, h=5,k=5 and r=5 .

Substitute the values in standard equation of circle,

  (x5)2+(y5)2=52(x5)2+(y5)2=25

Thus, the equation of circle is (x5)2+(y5)2=25 .

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