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. Can someone tell me what it means when it says "reflection of circle C in the y-axis"?
I can't interpret the image really. Thanks all.
circle y-axis circle reflection
Hi bonkysleuth,
The topic of equation of a circle is a new topic introduced in the new syllabus 2008 of Additional Mathematics.
The equation of the circle was once set in the late 60s and early 70s. If you need more practices, you can go to Bras Basah Complex second hand bookshop to look for the old, old ten years series.
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
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
A Practice Question
(a) Show that the line 4y = x - 3 touches the circle
x^2 + y^2 - 4x - 8y + 3 = 0
(b) The straight line x = 3 intersects the circle x^2 + y^2 - 8x - 10y - 9 = 0 at the
points P and Q. Find the equations of tangents at P and Q.
Regards,
ahm97sic
(x-2)^2 + (y+3)^2 = 5^2
reflecting about y-axies only change the x-value so
(x+2)^2 + (y+3)^2 = 5^2
Originally posted by weewee:(x-2)^2 + (y+3)^2 = 5^2
reflecting about y-axies only change the x-value so
(x+2)^2 + (y+3)^2 = 5^2
Hi Weewee,
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
Thank you for your kind attention.
Regards,
ahm97sic
Originally posted by ^tamago^:practice A levels maths C also can.
Hi Tamago,
As far as I know, there are only questions set on the sketching of the circles, eclipse, hyperbola etc but there are no questions set on the finding of the equation of the circle, its centre, radius, tangent etc.
I have all the "A" level Maths C exam questions from 1968 to 2007. Please kindly let me know which years where questions are set on the finding, not sketching, of the equation of the circle, its centre, radius, tangent etc.
Thank you for your kind attention.
Regards,
ahm97sic