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
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!
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
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 !!
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
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)
Originally posted by eagle: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 workingsi.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)
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
Originally posted by Garrick_3658: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)
Hi Eagle,
Thanks for the constructive comments and method. Your answer has been updated in the consolidated solution.
Regards,
ahm97sic