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DIFFERENTIAL CALCULUS
Study Notes 4

“CIRCLES”

Prepared By: Tutor Win

CIRCLES

07/29/2021

In the last StudyNote 3, we have learned the foundation of lines
through determining its steepness or slope, its equations, procedures on
how to get the equation and the like
...
In this note, we will
learn the foundation of circles and its equation
...


EQUATIONS OF CIRCLES
The standard equation of a circle may be associated to the distance
formula
...
The standard equation of the circle is
(𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟

A(x,y)
r

C(h,k)

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Example 1
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Step 1
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(𝑥 − 2) + (𝑦 − 3) = 5
Step 3
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Find the standard equation of the circle having a center at (3,-2)
and radius of 2 units
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(𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟
Step 2
...

(𝑥 − 3) + (𝑦 + 2) = 4
Example 3
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Step 1: (𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟
Step 2
...

Step 3
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If we have,
(x+6)2 + (y-2)2 = 49
It may also be equivalent to,
x2 + y2 + 12x – 4y – 9 = 0
This is the general equation of the circle
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We
can expand (x+6)2 using multiplication of binomial or FOIL method
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We can also expand (y-2)2
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(x+6)2 + (y-2)2 = 49
(x2 + 12x + 36) + (y2 – 4y + 4) = 49
Next is to evaluate by removing the parentheses and combining like
terms
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How can we now convert the general equation of a circle to its standard
equation? Let’s use again the example above
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Step 1
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(x2 + 12x) + (y2 – 4y) – 9 = 0
Now we have grouped them
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To do so,
(x2

Let (x2 + 12x) = (x2 + Ax); and
(y2 – 4y) = (y2 + By)
To find the constant term of each, solve for
( )2 (in the trinomial with x variable) and
( )2 (in the trinomial with y variable)

A = 12 and B = -4
( )2 = ( )2 = (6)2 = 36
( )2 = ( )2 = (−2)2 = 4

Therefore,
(x2 + 12x + 36) and (y2 – 4y + 4)

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(x2 + 12x + 36) + (y2 – 4y + 4) – 9 = 0 + 36 + 4
Note: We have just added 36 and 4 to the left side of the equation
...
That’s why
we have also added 36 and 4 to the right side of the equation
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(x2 + 12x + 36) + (y2 – 4y + 4) – 9 = 40
(x+ )2 + (y+ )2 – 9 = 40
(x+6)2 + (y-2)2 – 9 = 40
(x+6)2 + (y-2)2 – 9 + 9= 40 + 9
(x+𝟔)2 + (y-2)2 = 49

Example 4
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Step 1
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(x2 - 2x +1) + (y2 – 6y + 9) = 100

Step 3
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x2 + y2 - 2x – 6y +1 + 9 = 100

Step 5
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x2 + y2 - 2x – 6y +10 – 100 = 100 – 100

Step 7
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Convert x2 + y2 + 8x – 16y – 64 = 0 to its standard form
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x2 + y2 + 8x – 16y – 64 = 0

Step 2
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[x2 + 8x + ( )2] + [y2 – 16y + (

Step 4
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(x2 + 8x + 16) + (y2 – 16y + 64) – 64 = 80

Step 6
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(x + 4)2 + (y – 8)2 – 64 = 80

Step 8
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(x + 4)2 + (y – 8)2 = 144

)2] – 64 = 0 + ( )2 + (

)2

)2 – 64 = 80

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PRACTICE TEST 4
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1
...

3
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(9, 1)
r=4
(-1, 5)
r=6
(2, -2)
r = √7

4
...
(- , )

r = 2√5

B
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7
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Convert standard equation to general equation
...
Convert general equation to standard equation
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x2 + y2 + 6x – 4y – 3 = 0
10
...

x2 + y2 – 6x + 8y + 24 = 0

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CIRCLES

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PRACTICE TEST 4
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1
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(𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟
Step 2
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(𝑥 − 9) + (𝑦 − 1) = 16
2
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(𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟
Step 2
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(𝑥 + 1) + (𝑦 − 5) = 36
3
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(𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟
Step 2
...

(𝑥 − 2) + (𝑦 + 2) = 7
4
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(𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟
Step 2
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(𝑥 + 12) + (𝑦 + 3) =

1
4

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5
...

(𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟
Step 2
...

1
5
(𝑥 − ) + (𝑦 − ) = 20
3
6
B
...
Step 1
...

Step 3
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Step 5
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Step 7
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Step 1
...

Step 3
...

Step 5
...

Step 7
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Step 1
...

Step 3
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Step 5
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Step 7
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9
...
x2 + y2 + 6x – 4y – 3 = 0
Step 2
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[x2 + 6x + ( )2] + [y2 – 4y + ( )2] – 3 = 0 + ( )2 + ( )2
Step 4
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Step 6
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Step 8
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(x + 3)2 + (y – 2)2 –3 = 13

(x2 + 6x + 9) + (y2 – 4y + 4) – 3 = 13
(x + )2 + (y – )2 – 3 = 13
(x + 3)2 + (y – 2)2 – 3 + 3 = 13 + 3
(x + 3)2 + (y – 2)2 = 16

10
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x2 + y2 -2x – 4y – 4 = 0
Step 2
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[x2 - 2x + ( )2] + [y2 – 4y + ( )2] – 4 = 0 + ( )2 + ( )2
Step 4
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Step 6
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Step 8
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(x - 1)2 + (y – 2)2 – 4 = 5

(x2 - 2x + 1) + (y2 – 4y + 4) – 4 = 5
(x - )2 + (y – )2 – 4 = 5
(x - 1)2 + (y – 2)2 – 4 + 4 = 5 + 4
(x - 1)2 + (y – 2)2 = 9

11
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x2 + y2 - 6x + 8y + 24 = 0
Step 2
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[x2 - 6x + ( )2] + [y2 + 8y + ( )2] + 24 = 0 + ( )2 + ( )2
Step 4
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Step 6
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Step 8
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(x - 3)2 + (y – 4)2 + 24 = 25

(x2 - 6x + 9) + (y2 + 8y + 16) + 24 = 25
(x - )2 + (y – )2 + 24 = 25
(x - 3)2 + (y – 4)2 + 24 - 24 = 25 - 24
(x - 3)2 + (y – 4)2 = 1

10 | P a g e



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