Database - Equation of a Circle

Ahm97sic SG

Question 1

The equation of a circle is x^2 + y^2 - 4x + 6y - 12 =0

Find the equation of the new circle C1, which is a reflection of the circle C in the y-axis.

Answer

x^2 + y^2 - 4x + 6y - 12 = 0

x^2 + y^2 + 2x(-2) + 2y(3) - 12 = 0

Since the format of the equation of a circle is

x^2 + y^2 + 2gx + 2fy + c = 0

So, g = - 2 , f = 3, c= - 12

and the centre of a circle is ( -g, -f )

So, the centre = (-(-2), -3) = (2, - 3)

so, the radius = square root (g^2 + f^2 - c) = square root(2^2 +(-3)^2 - (-12)) = 5 units

Alternative method to find the centre and radius of the equation of a circle

x² + y² - 4x + 6y - 12 = 0
x² - 4x + y² + 6y - 12 = 0

x² - 4x + 4 - 4 + y² + 6y + 9 - 9 - 12 = 0
(x² - 4x + 4) + (y² + 6y + 9) - 4 - 9 - 12 = 0

(x² - 4x + 2^2) + (y² + 6y + 3^2) - 4 - 9 - 12 = 0

(x² - 4x + 2^2) + (y² + 6y + 3^2) - 25 = 0

(x -2)² + (y+3)² - 25 = 0
(x -2)² + (y+3)² = 25
(x -2)² + (y- (-3))² = 5²

Hence, the circle has radius 5 at centre (2, -3)

Since the new circle is a reflection in the y axis, then the centre of the new circle is

(-2, -3) and the radius is also 5 units.

So, the equation of the new circle is

(x - a)^2 + (y- b)^2 = r^2

where centre (a, b) and r is radius

Since the new centre is (-2, -3) and radius is 5 units, so the equation of the new circle is

(x - a)^2 + (y- b)^2 = r^2

(x -(-2))^2 + (y-(-3))^2 = 5^2

(x+2)^2 +(y+3)^2 = 25

x^2 +4x + 4 + y^2 +6y + 9 -25 = 0

x^2 + y^2 + 4x + 6y -12 = 0

Note : It is to be noted that reflected in the y axis,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 + 4x + 6y - 12 = 0

It is to be noted that reflected in the x axis,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 - 4x - 6y - 12 = 0

If the circle is reflected in the line y = x,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 + 6x - 4y - 12 = 0

If the circle is reflected in the line y = - x,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 - 6x + 4y - 12 = 0

Ahm97sic SG

Question

Find the equation of the circle which passes through A(8, 1) and B(7, 0) and has, for its tangent at B, the line 3x - 4y - 21 = 0

Answer

Step 1 : Rearrange the line 3x - 4y - 21 = 0 into the form of y = mx + c

4y = 3x - 21

y = (3/4)x -21/4

Hence, m = 3/4

Step 2 Find the gradient of BC where B(7,0) and centre C(-g, -f)

m1 = [0-(-f)]/[7-(-g)]

= f/[7+g]

Step 3 Since the line 3x - 4y - 21 = 0 is tangent to the circle at point B, so the line

3x - 4y = 0 is perpendicular to BC ie m = - 1/m1

3/4 = - (7+g)/f

f = - (4/3)(7 + g) ------------------ (0)

Step 4 Since the points A(8,1) and B(7,0) are points on the circle, substitute them

into the equation of the circle to form 2 simultaneous equations

Equation of the circle is x^2 + y^2 +2gx +2fy + c = 0

Substitute point A (8,1), 8^2 + 1^2 + 2g(8) + 2f(1) + c = 0

65 + 16g + 2f + c = 0 ------------- (1)

Substitute point B (7, 0), 7^2 + 0^2 + 2g(7) + 2f(0) + c = 0

49 + 14g + c = 0 ------------------ (2)

Equation (2) - Equation (1)

16 + 2g + 2f = 0

8 + g + f = 0

f = - 8 - g ---------------------------- (3)

Step 5 Susbtitute equation (3) into equation (0) to solve for g

Substitute f = - 8 - g into equation (0)

- 8 - g = - (4/3)(7 + g)

- 24 - 3g = - 28 - 4g

g = - 4

Step 6 Substitute g = - 4 into equation (3) to solve for f

f = - 8 - (- 4)

f = - 4

Step 7 Substitute g = - 4 into equation (2) to solve for c

49 + 14 (- 4) + c = 0

c = 7

Step 8 Substitue g = - 4 , f = - 4 and c = 7 into the equation of the circle

x^2 + y^2 +2gx +2fy + c = 0

x^2 + y^2 + 2(- 4)x + 2(- 4)y + 7 = 0

x^2 + y^2 - 8x - 8y + 7 = 0

Garrick_3658 SG

What if they asked,

Reflected in the line y = 1?

Distance between the two centres = y1 + 1 + y2 + 1 (y1 is the centre of initial circle, y2 is centre of reflected circle)

Am I correct to say so?

If it reflected in the line x = 1, I'll just have to change the "y" in the above equation to "x", right?

And of course, when I use "+1", it is just an example. It can be any integer.

Thanks!

Ahm97sic SG

Dear Darrick_3658,

If the circle is reflected in the line x = 1, you just have to change the x part in the equation of the circle as only x-ordinate is affected in the co-ordinates of centre of the new circle.

If the circle is reflected in the line y =1, you just have to change the y part in the equation of the circle as only y-ordinate is affected in the co-ordinates of the centre of the new circle.

Regards,

ahm97sic

davidche SG

wow this qn is tough. I really dont think they wld test too hard since this is new in the syllabus.

davidche SG

So, the equation of the new circle is

(x - a)^2 + (y- b)^2 = r^2

where centre (a, b) and r is radius

Since the new centre is (-2, -3) and radius is 5 units, so the equation of the new circle is

(x - a)^2 + (y- b)^2 = r^2

This part is repetitive. Actually in the exam i think you have to write the workings if it is more than 1 mark right?

Secondly, i dont quite understand wad happens to the equation of the reflected circle when it is reflected at line y=1

Thanks !!

Ahm97sic SG

wow this qn is tough. I really dont think they wld test too hard since this is new in the syllabus.

Dear Davidche,

The two questions that I posted in the thread " Must know new 2008 Add

Questions" are actual past "O" level add maths exam questions with slight

modification.

The questions on the equation of a circle (in the mid 60s and early 70s) in the past

"O" level add maths do test on tangent, passes through two points, circle touches

the axis and so on.

Perhaps, this is the first year, the "O" level examiners will test on the easier

question on the equation of a circle first.

Regards,

ahm97sic

davidche SG

Dear Davidche,

The two questions that I posted in the thread " Must know new 2008 Add

Questions" are actual past "O" level add maths exam questions with slight

modification.

The questions on the equation of a circle (in the mid 60s and early 70s) in the past

"O" level add maths do test on tangent, passes through two points, circle touches

the axis and so on.

Perhaps, this is the first year, the "O" level examiners will test on the easier

question on the equation of a circle first.

Regards,

ahm97sic

I am relieved that you said this XD

Ahm97sic SG

Dear Davidche,

(1) Since the centre of the original circle is (2, -3),

a reflection in the line x = 1,

the centre of the new circle will be (0, -3).

So, when we substitute (0, -3) = (a, b) into the equation of the circle

(x - a)^2 + (y- b)^2 = r^2

we only need to change the x part of the equation of the circle

(2) Since the centre of the original circle is (2, -3),

a reflection in the line y = 1,

the centre of the new circle will be (2, 5).

So, when we substitute (2, 5) = (a, b) into the equation of the circle

(x - a)^2 + (y- b)^2 = r^2

we only need to change the y part of the equation of the circle

Regards,

ahm97sic

Ahm97sic SG

Dear Davidche,

Keep calm and don't panic is most important when you are taking the exam.

Do those questions that you can do easily first and avoids carelessness, once you realise the numbers of questions that you can solve, you will be more confident and do well in th exam.

Even if the exam is difficult, don't panic. Do parts of the question that you can do and get some marks here and there, you can still pass the exam. So, keep calm and don't panic in any exam.

Have enough sleep, otherwise it will affect your performance during the exam.

All the best for your add maths exam tomorrow.

Regards,

ahm97sic

PS : Please come to my forums more often. OK.

eagle SG

I would rather make the equation nicer than memorise it with the g and f in it
===> reduces the number of things to memorise, and not much more workings

i.e.

x² + y² - 4x + 6y - 12 = 0
x² - 4x + y² + 6y - 12 = 0
(x² - 4x + 4) + (y² + 6y + 3²) -12 -4 -9 = 0 ==> sort of completing the square twice
(x -2)² + (y+3)² = 25
(x -2)² + (y+3)² = 5²

Hence, the circle has radius 5 at centre (2, -3)

davidche SG

I would rather make the equation nicer than memorise it with the g and f in it
===> reduces the number of things to memorise, and not much more workings

i.e.

Yup i agree. Just know (x -2)² + (y+3)² = 5² this formula can liao. Everything just change it to like that XD

And thanks Ahm97sic i totally understand liao XD

eagle SG

What if they asked,

Reflected in the line y = 1?

Distance between the two centres = y1 + 1 + y2 + 1 (y1 is the centre of initial circle, y2 is centre of reflected circle)

Am I correct to say so?

If it reflected in the line x = 1, I'll just have to change the "y" in the above equation to "x", right?

And of course, when I use "+1", it is just an example. It can be any integer.

Thanks!

Hmm...

One easy method to find the new centre if reflected across y=1 (horizontal line, means y coordinate change) would be

since the value if y = -3, the distance from y=-3 to y=1 is 4

So, the y value of the reflected centre is at 1+4 = 5

Hence, new centre is (2,5)

davidche SG

Haha i will. And yeah i sleep at mid night kinda late but I wake up at 9.30. Shld be ok lar. 9 hours.

Ahm97sic SG

Hi Eagle,

Thanks for the constructive comments and method. Your answer has been updated in the consolidated solution.

Regards,

ahm97sic

Ahm97sic SG

Dear Davidche,

Indeed, enough sleep is important for the best performance in exam.

All the best for your add maths exam.

Warmest Regards,

ahm97sic