SHOW THAT THE EQUATION DOES NOT REPRESENTS A CIRCLE

Problem 1 :

Explain why x² + y² + 6x - 4y + 17 = 0 looks like it represents a circle but, in fact does not.

Solution :

x² + y² + 6x - 4y + 17 = 0

Finding radius of circle.

Comparing with general form

2g = 6, g ==> 3

2f = -4, f ==> -2

c = 17

Radius = √g² + f² - c

Radius = √3² + (-2)² - 17

Radius = √(9 + 4 - 17)

Radius = √-4

Since the radius is unreal, the given equation will not represent the circle.

Problem 2 :

Explain why x² + y² + 2x + 3y + 5 = 0 looks like it represents a circle.

Solution :

x² + y² + 2x + 3y + 5 = 0

Finding radius of circle.

Comparing with general form

2g = 2, g ==> 1

2f = 3, f ==> 3/2

c = 5

Radius = √g² + f² - c

Radius = √1² + (3/2)² - 5

Radius = √(1 + (9/4) - 5)

Radius = √(13/4) - 5

Radius = √(-7/4)

Since the radius is unreal, the given equation will not represent the circle.

Problem 3 :

Show that the equation x² + y² - 3x + 3y + 10 = 0 does not represent the circle.

Solution :

x² + y² - 3x + 3y + 10 = 0

Finding radius of circle.

Comparing with general form

2g = -3, g ==> -3/2

2f = 3, f ==> 3/2

c = 10

Radius = √g² + f² - c

Radius = √(-3/2)² + (3/2)² - 10

Radius = √(9/4) + (9/4) - 10

Radius = √(18/4) - 10

Radius = √(-22/4)

= √(-11/2)

Since the radius is unreal, the given equation will not represent the circle.

Problem 4 :

Find the points where the line with equation y = 2x + 6 and circle with equation x² + y² + 2x + 2y - 8 = 0 intersect.

Solution :

Equation of line is y = 2x + 6

Equation of circle is x² + y² + 2x + 2y - 8 = 0

Applying the value of y in the equation of circle, we get

x² + (2x + 6)² + 2x + 2(2x + 6) - 8 = 0

x² + (2x)² + 2(2x)(6) + 6² + 2x + 4x + 12 - 8 = 0

x² + 4x² + 24x + 36 + 6x + 4 = 0

5x² + 30x + 40 = 0

x² + 6x + 8 = 0

(x + 2)(x + 4) = 0

x = -2 and x = -4

Applying x = -2, we get y = 2(-2) + 6

y = -4 + 6

y = 2

Applying x = -4, we get y = 2(-4) + 6

y = -8 + 6

y = -2

so, the points of intersections are (-2, -2) and (-4, -2).

Problem 5 :

Find the points where the line with equation x + y = 4 and circle with equation x² + y² + 6x + 2y - 22 = 0 intersect.

Solution :

Equation of line is x + y = 4

y = 4 - x

Equation of circle is x² + y² + 6x + 2y - 22 = 0

Applying the value of y in the equation of circle, we get

x² + (4 - x)² + 6x + 2(4 - x) - 22 = 0

x² + 4² - 2(4)x + x² + 6x + 8 - 2x - 22 = 0

2x² - 8x - 2x + 6x + 24 - 22 = 0

2x² - 4x + 2 = 0

x² - 2x + 1 = 0

(x - 1)(x - 1) = 0

x = 1 and x = 1

When x = 1, y = 4 - 1

y = 3

So, the point of intersection of the line and circle is at (1, 3). Then it must be the tangent line.

Problem 6 :

Prove that the circles x² + y² + 6x - 8y + 9 = 0 and x² + y² - 10x + 4y - 7 = 0 just touch each other at a single point.

Solution :

Since we are supposed to prove they are touching each other externally, we prove the distance between two radii = distance between the centers,

Center of the circle x² + y² + 6x - 8y + 9 = 0 :

x² + 6x + y² - 8y + 9 = 0

(x + 3)² - 3² + (y - 4)² - 4² + 9 = 0

(x + 3)² - 9 + (y - 4)² - 16 + 9 = 0

(x + 3)² + (y - 4)² = 16

Center (-3, 4), radius = 4

Center of the circle x² + y² - 10x + 4y - 7 = 0 :

x² - 10x + y² + 4y - 7 = 0

(x - 5)² - 5² + (y + 2)² - 2² - 7 = 0

(x - 5)² - 25 + (y + 2)² - 4 - 7 = 0

(x - 5)² + (y + 2)² - 36 = 0

(x - 5)² + (y + 2)²  = 36

Center (5, -2), radius = 6

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

= √(5 + 3)² + (-2 - 4)²

= √8² + (-6)²

= √(64+36)

= √100

= 10 ------(1)

4 + 6 = 10 ------(2)

So, the circles are touching each other at one point.