WRITE AN EQUATION OF THE CIRCLE WITH CENTER AND RADIUS

Problem 1 :

Find the standard form equation for the circle.

(i)  Center (1, 2) and radius 5.

(ii)  Center (-3, 2) and radius 1.

(iii)  Center (-1, -4) and radius 3.

(iv)  Center (0, 0) and radius √3.

Solution :

(i)  Center (1, 2) and radius 5.

Equation of the circle :

(x - h)² + (y - k)² = r²

(x - 1)² + (y - 2)² = 5²

(x-1)² + (y - 2)² = 25

(x - 1)² = x² - 2x + 1 and (y - 2)² = y² - 4y + 4

x² - 2x + 1 + y² - 4y + 4 = 25

x² + y² - 2x - 4y + 5 - 25 = 0

x² + y² - 2x - 4y - 20 = 0

So, the required equation of the circle is

x²+y²-2x-4y-20  =  0

(ii)  Center (-3, 2) and radius 1.

Solution :

Here h = -3 and k = 2

Equation of the circle :

(x - h )² + (y -  k)² = r²

(x +  3)² + (y -  2)² = 1²

(x + 3)² = x² + 6x + 9 and (y -  2)² = y²  - 4y + 4

x² + 6x + 9 + y²- 4y + 4 = 1

x² + y² + 6x - 4y + 9 + 4 - 1 = 0

x² + y² + 6x - 4y + 12 = 0

So, the required equation of the circle is

x² + y² + 6x - 4y + 12 = 0

(iii)  Center (-1, -4) and radius 3.

Solution :

(x - h)² + (y - k)² = r²

(x + 1)² + (y + 4)² = 3²

(x + 1)² = x² + 2x + 1 and (y + 4)² = y² + 8y + 16

x² + 2x + 1 + y² + 8y + 16 = 9

x² + y² + 2x + 8y + 17 - 9 = 0

x² + y² + 2x + 8y + 8 = 0

So, the required equation of the circle is

x² + y² + 2x + 8y + 8 = 0

(iv)  Center (0, 0) and radius √3.

Solution :

(x - h)² + (y - k)² = r²

(x - 0)² + (y - 0)² = √3²

x² + y² = 3

So, the required equation of the circle is

x² + y² = 3

Problem 2 :

A circle is graphed in the xy-plane. If the circle has a radius of 3 and the center of the circle is at (4, −2), which of the following could be an equation of the circle?

A) (x + 4)² + (y − 2)² = 3         B) (x + 4)² − (y − 2)² = 3

C) (x − 4)² + (y + 2)² = 9          D) (x − 4)² − (y + 2)² = 9

Solution :

Center is at (4, -2) and radius = 3

(x - h)² + (y − k)² = r²

(x - 4)² + (y − (-2))² = 3²

(x - 4)² + (y + 2)² = 9

So, option C is correct.

Problem 3 :

Which of the following equations is the equation of a circle that is centered at (2, −1) in the xy-plane and  passes through the point (6, −1)?

A) (x − 2)² + (y + 1)² = 4      B) (x + 2)² + (y − 1)² = 4

C) (x − 2)² + (y + 1)² = 16     D) (x + 2)² + (y − 1)² = 16

Solution :

To find radius of the circle, we need to find the distance between center and any point on the circle.

= √(x₂ - x₁)² + (y₂ - y₁)²

= √(6 - 2)² + (-1 + 1)²

= √4² + 0²

Radius = 4

r² = 4² ==> 16

(x - h)² + (y − k)² = r²

By applying center (2, −1) and r² = 16, we get

(x - 2)² + (y + 1)² = 16

So, option C is correct.

Problem 4 :

A circle has a radius of 10, and the x-coordinate of the center in the xy-plane is −2. Two points on the circumference of the circle are (6, 9) and (−8, −5). What is the y-coordinate of the center?

A) −3    B) 0     C) 3    D) 8

Solution :

Let the center be (-2, y)

Distance between center and any point on the circumference of the circle is radius.

radius = √(x₂ - x₁)² + (y₂ - y₁)²

10 = √(6 + 2)² + (9 - y)²

10² = 8² + (9 - y)²

100 - 64 = (9 - y)²

9 - y = √36

9 - y = 6 and 9 - y = -6

9 - 6 = y and 9 + 6 = y

y = 3 and y = 15

So, the required point as center be (-2, 3) and (-2, 15)

10 = √(-8 + 2)² + (-5 - 3)²

10 = √6² + 8²

10 = 10

10 = √(-8 + 2)² + (-5 - 15)²

10 = √6² + 20²

The value of y which is 15 is not satisfying the condition. So, the y-coordinate is 3.

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